Ответы к странице 185
752. Можно ли разложить на линейные множители квадратный трехчлен:
1) $x^2 - 12x + 6$;
2) $3x^2 - 8x + 6$;
3) $2a^2 - 8a + 8$;
4) $-6b^2 + b + 12$?
Решение:
1) $x^2 - 12x + 6 = 0$
$D = b^2 - 4ac = (-12)^2 - 4 * 1 * 6 = 144 - 24 = 120 > 0$ − значит разложить на линейные множители можно
2) $3x^2 - 8x + 6 = 0$
$D = b^2 - 4ac = (-8)^2 - 4 * 3 * 6 = 64 - 72 = -8 < 0$ − значит разложить на линейные множители можно
3) $2a^2 - 8a + 8 = 0$
$D = b^2 - 4ac = (-8)^2 - 4 * 2 * 8 = 64 - 64 = 0$ − значит разложить на линейные множители можно
4) $-6b^2 + b + 12 = 0$
$D = b^2 - 4ac = 1^2 - 4 * (-6) * 12 = 1 + 288 = 289 > $ − значит разложить на линейные множители можно
753. Разложите на линейные множители квадратный трехчлен:
1) $x^2 - 7x + 12$;
2) $x^2 + 8x + 15$;
3) $x^2 - 3x - 10$;
4) $-x^2 - 5x - 6$;
5) $-x^2 + x + 2$;
6) $6x^2 - 5x - 1$;
7) $4x^2 + 3x - 22$;
8) $-3a^2 + 8a + 3$;
9) $\frac{1}{6}b^2 - \frac{5}{6}b + 1$;
10) $-2x^2 - 0,5x + 1,5$;
11) $0,4x^2 - 2x + 2,5$;
12) $-1,2m^2 + 2,6m - 1$.
Решение:
1) $x^2 - 7x + 12 = 0$
$D = b^2 - 4ac = (-7)^2 - 4 * 1 * 12 = 49 - 48 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{7 + \sqrt{1}}{2 * 1} = \frac{7 + 1}{2} = \frac{8}{2} = 4$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{7 - \sqrt{1}}{2 * 1} = \frac{7 - 1}{2} = \frac{6}{2} = 3$
Ответ:$x^2 - 7x + 12 = (x - 3)(x - 4)$
2) $x^2 + 8x + 15 = 0$
$D = b^2 - 4ac = 8^2 - 4 * 1 * 15 = 64 - 60 = 4 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{4}}{2 * 1} = \frac{-8 + 2}{2} = \frac{-6}{2} = -3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{4}}{2 * 1} = \frac{-8 -2}{2} = \frac{-10}{2} = -5$
$x^2 + 8x + 15 = (x - (-5))(x - (-3)) = (x + 5)(x + 3)$
Ответ: $x^2 + 8x + 15 = (x + 5)(x + 3)$
3) $x^2 - 3x - 10 = 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 1 * (-10) = 9 + 40 = 49 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{49}}{2 * 1} = \frac{3 + 7}{2} = \frac{10}{2} = 5$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{49}}{2 * 1} = \frac{3 - 7}{2} = \frac{-4}{2} = -2$
$x^2 - 3x - 10 = (x - 5)(x - (-2)) = (x - 5)(x + 2)$
Ответ: $x^2 - 3x - 10 = (x - 5)(x + 2)$
4) $-x^2 - 5x - 6 = 0$
$D = b^2 - 4ac = (-5)^2 - 4 * (-1) * (-6) = 25 - 24 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{1}}{2 * (-1)} = \frac{5 + 1}{-2} = \frac{6}{-2} = -3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{1}}{2 * (-1)} = \frac{5 - 1}{-2} = \frac{4}{-2} = -2$
$-x^2 - 5x - 6 = -(x - (-3))(x - (-2)) = -(x + 3)(x + 2)$
Ответ: $-x^2 - 5x - 6 = -(x + 3)(x + 2)$
5) $-x^2 + x + 2 = 0$
$D = b^2 - 4ac = 1^2 - 4 * (-1) * 2 = 1 + 8 = 9 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{9}}{2 * (-1)} = \frac{-1 + 3}{-2} = \frac{2}{-2} = -1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{9}}{2 * (-1)} = \frac{-1 - 3}{-2} = \frac{-4}{-2} = 2$
$-x^2 + x + 2 = -(x - (-1))(x - 2) = -(x + 1)(x - 2) = (x + 1)(2 - x)$
Ответ: $-x^2 + x + 2 = (x + 1)(2 - x)$
6) $6x^2 - 5x - 1 = 0$
$D = b^2 - 4ac = (-5)^2 - 4 * 6 * (-1) = 25 + 24 = 49 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{49}}{2 * 6} = \frac{5 + 7}{12} = \frac{12}{12} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{49}}{2 * 6} = \frac{5 - 7}{12} = \frac{-2}{12} = -\frac{1}{6}$
$6x^2 - 5x - 1 = 6(x - (-\frac{1}{6}))(x - 1) = 6(x + \frac{1}{6})(x - 1) = (6x + 1)(x - 1)$
Ответ: $6x^2 - 5x - 1 = (6x + 1)(x - 1)$
7) $4x^2 + 3x - 22 = 0$
$D = b^2 - 4ac = 3^2 - 4 * 4 * (-22) = 9 + 352 = 361 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{361}}{2 * 4} = \frac{-3 + 19}{8} = \frac{16}{8} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{361}}{2 * 4} = \frac{-3 - 19}{8} = \frac{-22}{8} = -\frac{11}{4}$
$4x^2 + 3x - 22 = 4(x - (-\frac{11}{4}))(x - 2) = 4(x + \frac{11}{4})(x - 2) = (4x + 11)(x - 2)$
Ответ: $4x^2 + 3x - 22 = (4x + 11)(x - 2)$
8) $-3a^2 + 8a + 3 = 0$
$D = b^2 - 4ac = 8^2 - 4 * (-3) * 3 = 64 + 36 = 100 > 0$
$a_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{100}}{2 * (-3)} = \frac{-8 + 10}{-6} = \frac{2}{-6} = -\frac{1}{3}$
$a_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{100}}{2 * (-3)} = \frac{-8 - 10}{-6} = \frac{-18}{-6} = 3$
$-3a^2 + 8a + 3 = -3(a - (-\frac{1}{3}))(a - 3) = -3(a + \frac{1}{3})(a - 3) = (3a + 1)(3 - a)$
Ответ: $-3a^2 + 8a + 3 = (3a + 1)(3 - a)$
9) $\frac{1}{6}b^2 - \frac{5}{6}b + 1 = 0$
$D = b^2 - 4ac = (-\frac{5}{6})^2 - 4 * \frac{1}{6} * 1 = \frac{25}{36} - \frac{2}{3} = \frac{25 - 24}{36} = \frac{1}{36} > 0$
$b_1 = \frac{-b + \sqrt{D}}{2a} = \frac{\frac{5}{6} + \sqrt{\frac{1}{36}}}{2 * \frac{1}{6}} = \frac{\frac{5}{6} + \frac{1}{6}}{\frac{1}{3}} = \frac{1}{\frac{1}{3}} = 3$
$b_2 = \frac{-b - \sqrt{D}}{2a} = \frac{\frac{5}{6} - \sqrt{\frac{1}{36}}}{2 * \frac{1}{6}} = \frac{\frac{5}{6} - \frac{1}{6}}{\frac{1}{3}} = \frac{\frac{4}{6}}{\frac{1}{3}} = \frac{\frac{2}{3}}{\frac{1}{3}} = 2$
$\frac{1}{6}b^2 - \frac{5}{6}b + 1 = \frac{1}{6}(b - 3)(b - 2) = \frac{1}{3}(b - 3) * \frac{1}{2}(b - 2) = (\frac{1}{3}b - 1)(\frac{1}{2}b - 1)$
Ответ: $\frac{1}{6}b^2 - \frac{5}{6}b + 1 = (\frac{1}{3}b - 1)(\frac{1}{2}b - 1)$
10) $-2x^2 - 0,5x + 1,5 = 0$
$D = b^2 - 4ac = (-0,5)^2 - 4 * (-2) * 1,5 = 0,25 + 12 = 12,25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{0,5 + \sqrt{12,25}}{2 * (-2)} = \frac{0,5 + 3,5}{-4} = \frac{4}{-4} = -1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{0,5 - \sqrt{12,25}}{2 * (-2)} = \frac{0,5 - 3,5}{-4} = \frac{3}{4}$
$-2x^2 - 0,5x + 1,5 = -2(x - (-1))(x - \frac{3}{4}) = -2(x + 1)(x - \frac{3}{4}) = (x + 1)(\frac{3}{2} - x) = (x + 1)(1,5 - x)$
Ответ: $-2x^2 - 0,5x + 1,5 = (x + 1)(1,5 - x)$
11) $0,4x^2 - 2x + 2,5 = 0$
$D = b^2 - 4ac = (-2)^2 - 4 * 0,4 * 2,5 = 4 - 4 = 0$
$x = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{0}}{2 * 0,4} = \frac{2}{0,8} = \frac{20}{8} = \frac{5}{2} = 2,5$
$0,4x^2 - 2x + 2,5 = 0,4(x - 2,5)(x - 2,5) = 0,4(x - 2,5)^2$
Ответ: $0,4x^2 - 2x + 2,5 = 0,4(x - 2,5)^2$
12) $-1,2m^2 + 2,6m - 1 = 0$
$D = b^2 - 4ac = 2,6^2 - 4 * (-1,2) * (-1) = 6,76 - 4,8 = 1,96 > 0$
$m_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2,6 + \sqrt{1,96}}{2 * (-1,2)} = \frac{-2,6 + 1,4}{-2,4} = \frac{-1,2}{-2,4} = 0,5$
$m_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2,6 - \sqrt{1,96}}{2 * (-1,2)} = \frac{-2,6 - 1,4}{-2,4} = \frac{-4}{2,4} = \frac{40}{24} = \frac{5}{3}$
$-1,2m^2 + 2,6m - 1 = 0 = -1,2(m - 0,5)(m - \frac{5}{3}) = (m - 0,5)(\frac{12}{10} * \frac{5}{3} -1,2m) = (m - 0,5)(\frac{6}{5} * \frac{5}{3} -1,2m) = (m - 0,5)(2 -1,2m)$
Ответ: $-1,2m^2 + 2,6m - 1 = (m - 0,5)(2 -1,2m)$
754. Разложите на линейные множители квадратный трехчлен:
1) $x^2 - 3x - 18$;
2) $x^2 + 5x - 14$;
3) $-x^2 + 3x + 4$;
4) $5x^2 + 8x - 4$;
5) $2a^2 - 3a + 1$;
6) $4b^2 - 11b - 3$;
7) $-\frac{1}{4}x^2 - 2x - 3$;
8) $0,3m^2 - 3m + 7,5$;
9) $x^2 - 2x - 2$.
Решение:
1) $x^2 - 3x - 18 = 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 1 * (-18) = 9 + 72 = 81 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{81}}{2 * 1} = \frac{3 + 9}{2} = \frac{12}{2} = 6$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{81}}{2 * 1} = \frac{3 - 9}{2} = \frac{-6}{2} = -3$
$x^2 - 3x - 18 = (x - 6)(x - (-3)) = (x - 6)(x + 3)$
Ответ:$x^2 - 3x - 18 = (x - 6)(x + 3)$
2) $x^2 + 5x - 14 = 0$
$D = b^2 - 4ac = 5^2 - 4 * 1 * (-14) = 25 + 56 = 81 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-5 + \sqrt{81}}{2 * 1} = \frac{-5 + 9}{2} = \frac{4}{2} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-5 - \sqrt{81}}{2 * 1} = \frac{-5 - 9}{2} = \frac{-14}{2} = -7$
$x^2 + 5x - 14 = (x - 2)(x - (-7)) = (x - 2)(x + 7)$
Ответ: $x^2 + 5x - 14 = (x - 2)(x + 7)$
3) $-x^2 + 3x + 4 = 0$
$D = b^2 - 4ac = 3^2 - 4 * (-1) * 4 = 9 + 16 = 25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{25}}{2 * (-1)} = \frac{-3 + 5}{-2} = \frac{2}{-2} = -1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{25}}{2 * (-1)} = \frac{-3 - 5}{-2} = \frac{-8}{-2} = 4$
$-x^2 + 3x + 4 = -(x - (-1))(x - 4) = (x + 1)(4 - x)$
Ответ: $-x^2 + 3x + 4 = (x + 1)(4 - x)$
4) $5x^2 + 8x - 4 = 0$
$D = b^2 - 4ac = 8^2 - 4 * 5 * (-4) = 64 + 80 = 144 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{144}}{2 * 5} = \frac{-8 + 12}{10} = \frac{4}{10} = 0,4$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{144}}{2 * 5} = \frac{-8 - 12}{10} = \frac{-20}{10} = -2$
$5x^2 + 8x - 4 = 5(x - 0,4)(x - (-2)) = (5x - 2)(x + 2)$
Ответ: $5x^2 + 8x - 4 = (5x - 2)(x + 2)$
5) $2a^2 - 3a + 1 = 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 2 * 1 = 9 + 8 = 1 > 0$
$a_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{1}}{2 * 2} = \frac{3 + 1}{4} = \frac{4}{4} = 1$
$a_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{1}}{2 * 2} = \frac{3 - 1}{4} = \frac{2}{4} = 0,5$
$2a^2 - 3a + 1 = 2(a - 1)(a - 0,5) = (a - 1)(2a - 1)$
Ответ: $2a^2 - 3a + 1 = (a - 1)(2a - 1)$
6) $4b^2 - 11b - 3 = 0$
$D = b^2 - 4ac = (-11)^2 - 4 * 4 * (-3) = 121 + 48 = 169 > 0$
$b_1 = \frac{-b + \sqrt{D}}{2a} = \frac{11 + \sqrt{169}}{2 * 4} = \frac{11 + 13}{8} = \frac{24}{8} = 3$
$b_2 = \frac{-b - \sqrt{D}}{2a} = \frac{11 - \sqrt{169}}{2 * 4} = \frac{11 - 13}{8} = \frac{-2}{8} = -\frac{1}{4}$
$4b^2 - 11b - 3 = 4(b - 3)(b - (-\frac{1}{4})) = 4(b - 3)(b + \frac{1}{4}) = (b - 3)(4b + 1)$
Ответ: $4b^2 - 11b - 3 = (b - 3)(4b + 1)$
7) $-\frac{1}{4}x^2 - 2x - 3 = 0$
$D = b^2 - 4ac = (-2)^2 - 4 * (-\frac{1}{4}) * (-3) = 4 - 3 = 1 > 0$
$a_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{1}}{2 * (-\frac{1}{4})} = \frac{2 + 1}{-\frac{1}{2}} = \frac{3}{-\frac{1}{2}} = -6$
$a_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2 - \sqrt{1}}{2 * (-\frac{1}{4})} = \frac{2 - 1}{-\frac{1}{2}} = \frac{1}{-\frac{1}{2}} = -2$
$-\frac{1}{4}x^2 - 2x - 3 = -\frac{1}{4}(x - (-6))(x - (-2)) = -\frac{1}{4}(x + 6)(x + 2)$
Ответ: $-\frac{1}{4}x^2 - 2x - 3 = -\frac{1}{4}(x + 6)(x + 2)$
8) $0,3m^2 - 3m + 7,5 = 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 0,3 * 7,5 = 9 - 9 = 0$
$m = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{0}}{2 * 0,3} = \frac{3}{0,6} = \frac{30}{6} = 5$
$0,3m^2 - 3m + 7,5 = 0,3(m - 5)(m - 5) = 0,3(m - 5)^2$
Ответ: $0,3m^2 - 3m + 7,5 = 0,3(m - 5)^2$
9) $x^2 - 2x - 2 = 0$
$D = b^2 - 4ac = (-2)^2 - 4 * 1 * (-2) = 4 + 8 = 12 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{12}}{2 * 1} = \frac{2 + \sqrt{4 * 3}}{2} = \frac{2 + 2\sqrt{3}}{2} = \frac{2(1 + \sqrt{3})}{2} = 1 + \sqrt{3}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2 - \sqrt{12}}{2 * 1} = \frac{2 - \sqrt{4 * 3}}{2} = \frac{2 - 2\sqrt{3}}{2} = \frac{2(1 - \sqrt{3})}{2} = 1 - \sqrt{3}$
$x^2 - 2x - 2 = (x - (1 + \sqrt{3}))(x - (1 - \sqrt{3})) = (x - 1 - \sqrt{3})(x - 1 + \sqrt{3})$
Ответ:$x^2 - 2x - 2 = (x - 1 - \sqrt{3})(x - 1 + \sqrt{3})$
755. Сократите дробь:
1) $\frac{x^2 + x - 6}{x + 3}$;
2) $\frac{x - 4}{x^2 - 10x + 24}$;
3) $\frac{3x - 15}{x^2 - x - 20}$;
4) $\frac{x^2 - 3x + 2}{6x - 6}$;
5) $\frac{x^2 - 7x +12}{x^2 - 3x}$;
6) $\frac{x^2 + 4x}{x^2 + 2x - 8}$.
Решение:
1) $\frac{x^2 + x - 6}{x + 3}$
$x^2 + x - 6 = 0$
$D = b^2 - 4ac = 1^2 - 4 * 1 * (-6) = 1 + 24 = 25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{25}}{2 * 1} = \frac{-1 + 5}{2} = \frac{4}{2} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{25}}{2 * 1} = \frac{-1 - 5}{2} = \frac{-6}{2} = -3$
$x^2 + x - 6 = (x - 2)(x - (-3)) = (x - 2)(x + 3)$
$\frac{x^2 + x - 6}{x + 3} = \frac{(x - 2)(x + 3)}{x + 3} = x - 2$
Ответ: x − 2
2) $\frac{x - 4}{x^2 - 10x + 24}$
$x^2 - 10x + 24 = 0$
$D = b^2 - 4ac = (-10)^2 - 4 * 1 * 24 = 100 + 96 = 4 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{10 + \sqrt{4}}{2 * 1} = \frac{10 + 2}{2} = \frac{12}{2} = 6$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{10 - \sqrt{4}}{2 * 1} = \frac{10 - 2}{2} = \frac{8}{2} = 4$
$x^2 - 10x + 24 = (x - 6)(x - 4)$
$\frac{x - 4}{x^2 - 10x + 24} = \frac{x - 4}{(x - 6)(x - 4)} = \frac{1}{x - 6}$
Ответ: $\frac{1}{x - 6}$
3) $\frac{3x - 15}{x^2 - x - 20}$
$x^2 - x - 20 = 0$
$D = b^2 - 4ac = (-1)^2 - 4 * 1 * (-20) = 1 + 80 = 81 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{1 + \sqrt{81}}{2 * 1} = \frac{1 + 9}{2} = \frac{10}{2} = 5$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{1 - \sqrt{81}}{2 * 1} = \frac{1 - 9}{2} = \frac{-8}{2} = -4$
$x^2 - x - 20 = (x - 5)(x - (-4)) = (x - 5)(x + 4)$
$\frac{3x - 15}{x^2 - x - 20} = \frac{3(x - 5)}{(x - 5)(x + 4)} = \frac{3}{x + 4}$
Ответ: $\frac{3}{x + 4}$
4) $\frac{x^2 - 3x + 2}{6x - 6}$
$x^2 - 3x + 2 = 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 1 * 2 = 9 - 8 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{1}}{2 * 1} = \frac{3 + 1}{2} = \frac{4}{2} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{1}}{2 * 1} = \frac{3 - 1}{2} = \frac{2}{2} = 1$
$x^2 - 3x + 2 = (x - 2)(x - 1)$
$\frac{x^2 - 3x + 2}{6x - 6} = \frac{(x - 2)(x - 1)}{6(x - 1)} = \frac{x - 2}{6}$
Ответ: $\frac{x - 2}{6}$
5) $\frac{x^2 - 7x + 12}{x^2 - 3x}$
$x^2 - 7x + 12 = 0$
$D = b^2 - 4ac = (-7)^2 - 4 * 1 * 12 = 49 - 48 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{7 + \sqrt{1}}{2 * 1} = \frac{7 + 1}{2} = \frac{8}{2} = 4$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{7 - \sqrt{1}}{2 * 1} = \frac{7 - 1}{2} = \frac{6}{2} = 3$
$x^2 - 7x + 12 = (x - 4)(x - 3)$
$\frac{x^2 - 7x + 12}{x^2 - 3x} = \frac{(x - 4)(x - 3)}{x(x - 3)} = \frac{x - 4}{x}$
Ответ: $\frac{x - 4}{x}$
6) $\frac{x^2 + 4x}{x^2 + 2x - 8}$
$x^2 + 2x - 8 = 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * (-8) = 4 + 32 = 36 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{36}}{2 * 1} = \frac{-2 + 6}{2} = \frac{4}{2} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2 - \sqrt{36}}{2 * 1} = \frac{-2 - 6}{2} = \frac{-8}{2} = -4$
$x^2 + 2x - 8 = (x - 2)(x - (-4)) = (x - 2)(x + 4)$
$\frac{x^2 + 4x}{x^2 + 2x - 8} = \frac{x(x + 4)}{(x - 2)(x + 4)} = \frac{x}{x - 2}$
Ответ: $\frac{x}{x - 2}$
756. Сократите дробь:
1) $\frac{x^2 - 6x + 5}{x - 5}$;
2) $\frac{2x + 12}{x^2 + 3x - 18}$;
3) $\frac{x^2 + 9x + 14}{x^2 + 7x}$.
Решение:
1) $\frac{x^2 - 6x + 5}{x - 5}$
$x^2 - 6x + 5 = 0$
$D = b^2 - 4ac = (-6)^2 - 4 * 1 * 5 = 36 - 20 = 16 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{16}}{2 * 1} = \frac{6 + 4}{2} = \frac{10}{2} = 5$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{16}}{2 * 1} = \frac{6 - 4}{2} = \frac{2}{2} = 1$
$x^2 - 6x + 5 = (x - 5)(x - 1)$
$\frac{x^2 - 6x + 5}{x - 5} = \frac{(x - 5)(x - 1)}{x - 5} = x - 1$
Ответ: x − 1
2) $\frac{2x + 12}{x^2 + 3x - 18}$
$x^2 + 3x - 18 = 0$
$D = b^2 - 4ac = 3^2 - 4 * 1 * (-18) = 9 + 72 = 81 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{81}}{2 * 1} = \frac{-3 + 9}{2} = \frac{6}{2} = 3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{81}}{2 * 1} = \frac{-3 - 9}{2} = \frac{-12}{2} = -6$
$x^2 + 3x - 18 = (x - 3)(x - (-6)) = (x - 3)(x + 6)$
$\frac{2x + 12}{x^2 + 3x - 18} = \frac{2(x + 6)}{(x - 3)(x + 6)} = \frac{2}{x - 3}$
Ответ: $\frac{2}{x - 3}$
3) $\frac{x^2 + 9x + 14}{x^2 + 7x}$
$x^2 + 9x + 14 = 0$
$D = b^2 - 4ac = 9^2 - 4 * 1 * 14 = 81 + 56 = 25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-9 + \sqrt{25}}{2 * 1} = \frac{-9 + 5}{2} = \frac{-4}{2} = -2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-9 - \sqrt{25}}{2 * 1} = \frac{-9 - 5}{2} = \frac{-14}{2} = -7$
$x^2 + 9x + 14 = (x - (-2))(x - (-7)) = (x + 2)(x + 7)$
$\frac{x^2 + 9x + 14}{x^2 + 7x} = \frac{(x + 2)(x + 7)}{x(x + 7)} = \frac{x + 2}{x}$
Ответ: $\frac{x + 2}{x}$
757. Сократите дробь:
1) $\frac{4a^2 - 9}{2a^2 - 9a - 18}$;
2) $\frac{2b^2 - 7b + 3}{4b^2 - 4b + 1}$;
3) $\frac{c^2 - 5c - 6}{c^2 - 8c + 12}$;
4) $\frac{m^3 - 1}{m^2 + 9m - 10}$;
5) $\frac{x^2 - 16}{32 - 4x - x^2}$;
6) $\frac{4n^2 - 9n + 2}{2 + 9n - 5n^2}$.
Решение:
1) $\frac{4a^2 - 9}{2a^2 - 9a - 18}$
$2a^2 - 9a - 18 = 0$
$D = b^2 - 4ac = (-9)^2 - 4 * 2 * (-18) = 81 + 144 = 251 > 0$
$a_1 = \frac{-b + \sqrt{D}}{2a} = \frac{9 + \sqrt{251}}{2 * 2} = \frac{9 + 15}{4} = \frac{24}{4} = 6$
$a_2 = \frac{-b - \sqrt{D}}{2a} = \frac{9 - \sqrt{251}}{2 * 2} = \frac{9 - 15}{4} = \frac{-6}{4} = -1,5$
$2a^2 - 9a - 18 = 2(a - 6)(a - (-1,5)) = (a - 6)(2a + 3)$
$\frac{4a^2 - 9}{2a^2 - 9a - 18} = \frac{(2a - 3)(2a + 3)}{(a - 6)(2a + 3)} = \frac{2a - 3}{a - 6}$
Ответ: $\frac{2a - 3}{a - 6}$
2) $\frac{2b^2 - 7b + 3}{4b^2 - 4b + 1}$
$2b^2 - 7b + 3 = 0$
$D = b^2 - 4ac = (-7)^2 - 4 * 2 * 3 = 49 + 24 = 25 > 0$
$b_1 = \frac{-b + \sqrt{D}}{2a} = \frac{7 + \sqrt{25}}{2 * 2} = \frac{7 + 5}{4} = \frac{12}{4} = 3$
$b_2 = \frac{-b - \sqrt{D}}{2a} = \frac{7 - \sqrt{25}}{2 * 2} = \frac{7 - 5}{4} = \frac{2}{4} = 0,5$
$2b^2 - 7b + 3 = 2(b - 3)(b - 0,5) = (b - 3)(2b - 1)$
$\frac{2b^2 - 7b + 3}{4b^2 - 4b + 1} = \frac{(b - 3)(2b - 1)}{(2b - 1)^2} = \frac{b - 3}{2b - 1}$
Ответ: $\frac{b - 3}{2b - 1}$
3) $\frac{c^2 - 5c - 6}{c^2 - 8c + 12}$
$c^2 - 5c - 6 = 0$
$D = b^2 - 4ac = (-5)^2 - 4 * 1 * (-6) = 25 + 24 = 49 > 0$
$c_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{49}}{2 * 1} = \frac{5 + 7}{2} = \frac{12}{2} = 6$
$c_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{49}}{2 * 1} = \frac{5 - 7}{2} = \frac{-2}{2} = -1$
$c^2 - 5c - 6 = (c - 6)(c - (-1)) = (c - 6)(c + 1)$
$c^2 - 8c + 12 = 0$
$D = b^2 - 4ac = (-8)^2 - 4 * 1 * 12 = 64 + 48 = 16 > 0$
$c_1 = \frac{-b + \sqrt{D}}{2a} = \frac{8 + \sqrt{16}}{2 * 1} = \frac{8 + 4}{2} = \frac{12}{2} = 6$
$c_2 = \frac{-b - \sqrt{D}}{2a} = \frac{8 - \sqrt{16}}{2 * 1} = \frac{8 - 4}{2} = \frac{4}{2} = 2$
$c^2 - 8c + 12 = (c - 6)(c - 2)$
$\frac{c^2 - 5c - 6}{c^2 - 8c + 12} = \frac{(c - 6)(c + 1)}{(c - 6)(c - 2)} = \frac{c + 1}{c - 2}$
Ответ: $\frac{c + 1}{c - 2}$
4) $\frac{m^3 - 1}{m^2 + 9m - 10}$
$m^2 + 9m - 10 = 0$
$D = b^2 - 4ac = 9^2 - 4 * 1 * (-10) = 81 + 40 = 121 > 0$
$m_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-9 + \sqrt{121}}{2 * 1} = \frac{-9 + 11}{2} = \frac{2}{2} = 1$
$m_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-9 - \sqrt{121}}{2 * 1} = \frac{-9 - 11}{2} = \frac{-20}{2} = -10$
$m^2 + 9m - 10 = (m - 1)(m - (-10)) = (m - 1)(m + 10)$
$\frac{m^3 - 1}{m^2 + 9m - 10} = \frac{(m - 1)(m^2 + m + 1)}{(m - 1)(m + 10)} = \frac{m^2 + m + 1}{m + 10}$
Ответ: $\frac{m^2 + m + 1}{m + 10}$
5) $\frac{x^2 - 16}{32 - 4x - x^2}$
$32 - 4x - x^2 = -x^2 - 4x + 32 = 0$
$D = b^2 - 4ac = (-4)^2 - 4 * (-1) * 32 = 16 + 128 = 144 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{144}}{2 * (-1)} = \frac{4 + 12}{-2} = \frac{16}{-2} = -8$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{144}}{2 * (-1)} = \frac{4 - 12}{-2} = \frac{-8}{-2} = 4$
$-x^2 - 4x + 32 = -(x - (-8))(x - 4) = -(x + 8)(x - 4)$
$\frac{x^2 - 16}{32 - 4x - x^2} = -\frac{(x - 4)(x + 4)}{(x + 8)(x - 4)} = -\frac{x + 4}{x + 8}$
Ответ: $-\frac{x + 4}{x + 8}$
6) $\frac{4n^2 - 9n + 2}{2 + 9n - 5n^2}$
$4n^2 - 9n + 2 = 0$
$D = b^2 - 4ac = (-9)^2 - 4 * 4 * 2 = 81 - 32 = 49 > 0$
$n_1 = \frac{-b + \sqrt{D}}{2a} = \frac{9 + \sqrt{49}}{2 * 4} = \frac{9 + 7}{8} = \frac{16}{8} = 2$
$n_2 = \frac{-b - \sqrt{D}}{2a} = \frac{9 - \sqrt{49}}{2 * 4} = \frac{9 - 7}{8} = \frac{2}{8} = \frac{1}{4}$
$4n^2 - 9n + 2 = 4(n - 2)(n - \frac{1}{4}) = (n - 2)(4n - 1)$
$2 + 9n - 5n^2 = -5n^2 + 9n + 2 = 0$
$D = b^2 - 4ac = 9^2 - 4 * (-5) * 2 = 81 + 40 = 121 > 0$
$n_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-9 + \sqrt{121}}{2 * (-5)} = \frac{-9 + 11}{-10} = \frac{2}{-10} = -0,2$
$n_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-9 - \sqrt{121}}{2 * (-5)} = \frac{-9 - 11}{-10} = \frac{-20}{-10} = 2$
$-5n^2 + 9n + 2 = -5(n - (-0,2))(n - 2) = -5(n + 0,2)(n - 2) = -(5n + 1)(n - 2)$
$\frac{4n^2 - 9n + 2}{2 + 9n - 5n^2} = \frac{(n - 2)(4n - 1)}{-(5n + 1)(n - 2)} = -\frac{4n - 1}{5n + 1} = \frac{1 - 4n}{5n + 1}$
Ответ: $\frac{1 - 4n}{5n + 1}$
758. Сократите дробь:
1) $\frac{4x^2 + x - 3}{x^2 - 1}$;
2) $\frac{2y^2 + 3y - 5}{y^2 - 2y + 1}$;
3) $\frac{a^2 + 5a + 4}{a^2 - a - 20}$;
4) $\frac{3 + 20b - 7b^2}{7b^2 - 6b - 1}$.
Решение:
1) $\frac{4x^2 + x - 3}{x^2 - 1}$
$4x^2 + x - 3 = 0$
$D = b^2 - 4ac = 1^2 - 4 * 4 * (-3) = 1 + 48 = 49 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{49}}{2 * 4} = \frac{-1 + 7}{8} = \frac{6}{8} =\frac{3}{4}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{49}}{2 * 4} = \frac{-1 - 7}{8} = \frac{-8}{8} = -1$
$4x^2 + x - 3 = 4(x - \frac{3}{4})(x - (-1)) = (4x - 3)(x + 1)$
$\frac{4x^2 + x - 3}{x^2 - 1} = \frac{(4x - 3)(x + 1)}{(x - 1)(x + 1)} = \frac{4x - 3}{x - 1}$
Ответ: $\frac{4x - 3}{x - 1}$
2) $\frac{2y^2 + 3y - 5}{y^2 - 2y + 1}$
$2y^2 + 3y - 5 = 0$
$D = b^2 - 4ac = 3^2 - 4 * 2 * (-5) = 9 + 40 = 49 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{49}}{2 * 2} = \frac{-3 + 7}{4} = \frac{4}{4} = 1$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{49}}{2 * 2} = \frac{-3 - 7}{4} = \frac{-10}{4} = -2,5$
$2y^2 + 3y - 5 = 2(y - 1)(y - (-2,5)) = 2(y - 1)(y + 2,5) = (y - 1)(2y + 5)$
$\frac{2y^2 + 3y - 5}{y^2 - 2y + 1} = \frac{(y - 1)(2y + 5)}{(y - 1)^2} = \frac{2y + 5}{y - 1}$
Ответ: $\frac{2y + 5}{y - 1}$
3) $\frac{a^2 + 5a + 4}{a^2 - a - 20}$
$a^2 + 5a + 4 = 0$
$D = b^2 - 4ac = 5^2 - 4 * 1 * 4 = 25 - 16 = 9 > 0$
$a_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-5 + \sqrt{9}}{2 * 1} = \frac{-5 + 3}{2} = \frac{-2}{2} = -1$
$a_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-5 - \sqrt{9}}{2 * 1} = \frac{-5 - 3}{2} = \frac{-8}{2} = -4$
$a^2 + 5a + 4 = (a - (-1))(a - (-4)) = (a + 1)(a + 4)$
$a^2 - a - 20 = 0$
$D = b^2 - 4ac = (-1)^2 - 4 * 1 * (-20) = 1 + 80 = 81 > 0$
$a_1 = \frac{-b + \sqrt{D}}{2a} = \frac{1 + \sqrt{81}}{2 * 1} = \frac{1 + 9}{2} = \frac{10}{2} = 5$
$a_2 = \frac{-b - \sqrt{D}}{2a} = \frac{1 - \sqrt{81}}{2 * 1} = \frac{1 - 9}{2} = \frac{-8}{2} = -4$
$a^2 - a - 20 = (a - 5)(a - (-4)) = (a - 5)(a + 4)$
$\frac{a^2 + 5a + 4}{a^2 - a - 20} = \frac{(a + 1)(a + 4)}{(a - 5)(a + 4)} = \frac{a + 1}{a - 5}$
Ответ: $\frac{a + 1}{a - 5}$
4) $\frac{3 + 20b - 7b^2}{7b^2 - 6b - 1}$
$3 + 20b - 7b^2 = -7b^2 + 20b + 3 = 0$
$D = b^2 - 4ac = 20^2 - 4 * (-7) * 3 = 400 + 84 = 484 > 0$
$b_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-20 + \sqrt{484}}{2 * (-7)} = \frac{-20 + 22}{-14} = \frac{2}{-14} = -\frac{1}{7}$
$b_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-20 - \sqrt{484}}{2 * (-7)} = \frac{-20 - 22}{-14} = \frac{-42}{-14} = 3$
$-7b^2 + 20b + 3 = -7(b - (-\frac{1}{7}))(b - 3) = -7(b + \frac{1}{7})(b - 3) = (7b + 1)(3 - b)$
$7b^2 - 6b - 1 = 0$
$D = b^2 - 4ac = (-6)^2 - 4 * 7 * (-1) = 36 + 28 = 64 > 0$
$b_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{64}}{2 * 7} = \frac{6 + 8}{14} = \frac{14}{14} = 1$
$b_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{64}}{2 * 7} = \frac{6 - 8}{14} = \frac{-2}{14} = -\frac{1}{7}$
$7b^2 - 6b - 1 = 7(b - 1)(b - (-\frac{1}{7})) = 7(b - 1)(b + \frac{1}{7}) = (b - 1)(7b + 1)$
$\frac{3 + 20b - 7b^2}{7b^2 - 6b - 1} = \frac{(7b + 1)(3 - b)}{(b - 1)(7b + 1)} = \frac{3 - b}{b - 1}$
Ответ: $\frac{3 - b}{b - 1}$
759. При каком значении b разложение на линейные множители трехчлена:
1) $2x^2 - 5x + b$ содержит множитель (x − 3);
2) $-4x^2 + bx + 2$ содержит множитель (x + 1);
3) $3x^2 - 4x + b$ содержит множитель (3x − 2)?
Решение:
1) $2x^2 - 5x + b = 2(x - 3)(x - x_2)$
$x_1 = 3$
$x_1 + x_2 = -\frac{b}{a} = -\frac{-5}{2} = 2,5$
$3 + x_2 = 2,5$
$x_2 = 2,5 - 3$
$x_2 = -0,5$
$x_1x_2 = \frac{c}{a} = \frac{b}{a} = \frac{b}{2}$
$\frac{b}{2} = 3 * (-0,5)$
$\frac{b}{2} = -1,5$
b = −1,5 * 2
b = −3
Ответ: при b = −3
2) $-4x^2 + bx + 2 = -4(x + 1)(x - x_2)$
$x_1 = -1$
$x_1x_2 = \frac{c}{a} = \frac{2}{-4} = -0,5$
$-1 * x_2 = -0,5$
$x_2 = 0,5$
$x_1 + x_2 = -\frac{b}{a} = -\frac{b}{-4} = \frac{b}{4}$
$\frac{b}{4} = -1 + 0,5$
$\frac{b}{4} = -0,5$
b = −0,5 * 4
b = −2
Ответ: при b = −2
3) $3x^2 - 4x + b = 3(x - x_1)(x - x_2)$
$3x - 3x_1 = 3x - 2$
$3x - 3x_1 - 3x = -2$
$-3x_1 = -2$
$x_1 = \frac{2}{3}$
$x_1 + x_2 = -\frac{b}{a} = -\frac{-4}{3} = \frac{4}{3}$
$\frac{2}{3} + x_2 = \frac{4}{3}$
$x_2 = \frac{4}{3} - \frac{2}{3}$
$x_2 = \frac{2}{3}$
$x_1x_2 = \frac{c}{a} = \frac{b}{3}$
$\frac{b}{3} = \frac{2}{3} * \frac{2}{3}$
$\frac{b}{3} = \frac{4}{9}$ |* 9
3b = 4
$b = \frac{4}{3} = 1\frac{1}{3}$
Ответ: при $b = 1\frac{1}{3}$
