Ответы к странице 140
549. Вынесите множитель из−под знака корня:
1) $\sqrt{18x^{12}}$;
2) $\sqrt{y^9}$.
Решение:
1) $\sqrt{18x^{12}} = \sqrt{9 * 2 * (x^6)^2} = \sqrt{9} * \sqrt{2} * \sqrt{(x^6)^2} = 3\sqrt{2}x^6$
2) $\sqrt{y^9} = \sqrt{y^8 * y} = \sqrt{(y^4)^2 * y} = \sqrt{(y^4)^2} * \sqrt{y} = y^4\sqrt{y}$
550. Упростите выражение:
1) $\sqrt{98} - \sqrt{50} + \sqrt{32}$;
2) $3\sqrt{8} + \sqrt{128} - \frac{1}{3}\sqrt{162}$;
3) $0,7\sqrt{300} - 7\sqrt{\frac{3}{49}} + \frac{2}{3}\sqrt{108}$;
4) $\sqrt{5a} - 2\sqrt{20a} + 3\sqrt{80a}$;
5) $\sqrt{a^3b} - \frac{2}{a}\sqrt{a^5b}$, если a > 0;
6) $\sqrt{c^5} + 4c\sqrt{c^3} - 5c^2\sqrt{c}$.
Решение:
1) $\sqrt{98} - \sqrt{50} + \sqrt{32} = \sqrt{49 * 2} - \sqrt{25 * 2} + \sqrt{16 * 2} = 7\sqrt{2} - 5\sqrt{2} + 4\sqrt{2} = 6\sqrt{2}$
2) $3\sqrt{8} + \sqrt{128} - \frac{1}{3}\sqrt{162} = 3\sqrt{4 * 2} + \sqrt{64 * 2} - \frac{1}{3}\sqrt{81 * 2} = 3 * 2\sqrt{2} + 8\sqrt{2} - \frac{1}{3} * 9\sqrt{2} = 6\sqrt{2} + 8\sqrt{2} - 3\sqrt{2} = 11\sqrt{2}$
3) $0,7\sqrt{300} - 7\sqrt{\frac{3}{49}} + \frac{2}{3}\sqrt{108} = 0,7\sqrt{100 * 3} - 7\sqrt{\frac{1}{49} * 3} + \frac{2}{3}\sqrt{36 * 3} = 0,7 * 10\sqrt{3} - 7 * \frac{1}{7}\sqrt{3} + \frac{2}{3} * 6\sqrt{3} = 7\sqrt{3} - \sqrt{3} + 2 * 2\sqrt{3} = 6\sqrt{3} + 4\sqrt{3} = 10\sqrt{3}$
4) $\sqrt{5a} - 2\sqrt{20a} + 3\sqrt{80a} = \sqrt{5a} - 2\sqrt{4 * 5a} + 3\sqrt{16 * 5a} = \sqrt{5a} - 2 * 2\sqrt{5a} + 3 * 4\sqrt{5a} = \sqrt{5a} - 4\sqrt{5a} + 12\sqrt{5a} = 9\sqrt{5a}$
5) $\sqrt{a^3b} - \frac{2}{a}\sqrt{a^5b} = \sqrt{a^2 * ab} - \frac{2}{a}\sqrt{a^4 * ab} = a\sqrt{ab} - \frac{2}{a} * a^2\sqrt{ab} = a\sqrt{ab} - 2a\sqrt{ab} = -a\sqrt{ab}$, если a > 0
6) $\sqrt{c^5} + 4c\sqrt{c^3} - 5c^2\sqrt{c} = \sqrt{c^4 * c} + 4c\sqrt{c^2 * c} - 5c^2\sqrt{c} = c^2\sqrt{c} + 4c * c\sqrt{c} - 5c^2\sqrt{c} = c^2\sqrt{c} + 4c^2\sqrt{c} - 5c^2\sqrt{c} = 0$
551. Упростите выражение:
1) $0,5\sqrt{12} - 3\sqrt{27} + 0,4\sqrt{75}$;
2) $2,5\sqrt{28b} + \frac{2}{3}\sqrt{63b} - 10\sqrt{0,07b}$;
3) $\sqrt{81a^7} - 5a^3\sqrt{a} + \frac{6}{a}\sqrt{a^9}$.
Решение:
1) $0,5\sqrt{12} - 3\sqrt{27} + 0,4\sqrt{75} = 0,5\sqrt{4 * 3} - 3\sqrt{9 * 3} + 0,4\sqrt{25 * 3} = 0,5 * 2\sqrt{3} - 3 * 3\sqrt{3} + 0,4 * 5\sqrt{3} = \sqrt{3} - 9\sqrt{3} + 2\sqrt{3} = -6\sqrt{3}$
2) $2,5\sqrt{28b} + \frac{2}{3}\sqrt{63b} - 10\sqrt{0,07b} = 2,5\sqrt{4 * 7b} + \frac{2}{3}\sqrt{9 * 7b} - 10\sqrt{0,01 * 7b} = 2,5 * 2\sqrt{7b} + \frac{2}{3} * 3\sqrt{7b} - 10 * 0,1\sqrt{7b} = 5\sqrt{7b} + 2\sqrt{7b} - \sqrt{7b} = 6\sqrt{7b}$
3) $\sqrt{81a^7} - 5a^3\sqrt{a} + \frac{6}{a}\sqrt{a^9} = \sqrt{81a^6 * a} - 5a^3\sqrt{a} + \frac{6}{a}\sqrt{a^8 * a} = \sqrt{81 * (a^3)^2 * a} - 5a^3\sqrt{a} + \frac{6}{a}\sqrt{(a^4)^2 * a} = 9a^3\sqrt{a} - 5a^3\sqrt{a} + \frac{6}{a} * a^4\sqrt{a} = 4a^3\sqrt{a} + 6a^3\sqrt{a} = 10a^3\sqrt{a}$
552. Докажите, что:
1) $\sqrt{11 + 4\sqrt{7}} = \sqrt{7} + 2$;
2) $\sqrt{14 + 8\sqrt{3}} = \sqrt{8} + \sqrt{6}$.
Решение:
1) $\sqrt{11 + 4\sqrt{7}} = \sqrt{7 + 4 + 4\sqrt{7}} = \sqrt{(\sqrt{7})^2 + 2 * 2\sqrt{7} + 2^2} = \sqrt{(\sqrt{7} + 2)^2} = |\sqrt{7} + 2| = \sqrt{7} + 2$
2) $\sqrt{14 + 8\sqrt{3}} = \sqrt{8 + 6 + 2 * 4\sqrt{3}} = \sqrt{(\sqrt{8})^2 + 2\sqrt{16 * 3} + (\sqrt{6})^2} = \sqrt{(\sqrt{8})^2 + 2\sqrt{48} + (\sqrt{6})^2} = \sqrt{(\sqrt{8})^2 + 2\sqrt{8 * 6} + (\sqrt{6})^2} = \sqrt{(\sqrt{8} + \sqrt{6})^2} = |\sqrt{8} + \sqrt{6}| = \sqrt{8} + \sqrt{6}$
553. Упростите выражение:
1) $(2\sqrt{3} - 1)(\sqrt{27} + 2)$;
2) $(\sqrt{5} - 2)^2 - (3 + \sqrt{5})^2$;
3) $\sqrt{\sqrt{17} - 4} * \sqrt{\sqrt{17} + 4}$;
4) $(7 + 4\sqrt{3})(2 - \sqrt{3})^2$;
5) $(\sqrt{6 + 2\sqrt{5}} - \sqrt{6 - 2\sqrt{5}})^2$.
Решение:
1) $(2\sqrt{3} - 1)(\sqrt{27} + 2) = (2\sqrt{3} - 1)(\sqrt{9 * 3} + 2) = (2\sqrt{3} - 1)(3\sqrt{3} + 2) = 2\sqrt{3} * 3\sqrt{3} + 2\sqrt{3} * 2 - 1 * 3\sqrt{3} + (-1) * 2 = 6 * 3 + 4\sqrt{3} - 3\sqrt{3} - 2 = 18 + \sqrt{3} - 2 = 16 + \sqrt{3}$
2) $(\sqrt{5} - 2)^2 - (3 + \sqrt{5})^2 = (\sqrt{5})^2 - 2 * 2\sqrt{5} + 2^2) - (3^2 + 3 * 2\sqrt{5} + (\sqrt{5})^2) = 5 - 4\sqrt{5} + 4 - (9 + 6\sqrt{5} + 5) = 9 - 4\sqrt{5} - (14 + 6\sqrt{5}) = 9 - 4\sqrt{5} - 14 - 6\sqrt{5} = -5 - 10\sqrt{5}$
3) $\sqrt{\sqrt{17} - 4} * \sqrt{\sqrt{17} + 4} = \sqrt{(\sqrt{17} - 4)((\sqrt{17} + 4))} = \sqrt{(\sqrt{17})^2 - 4^2} = \sqrt{17 - 16} = \sqrt{1} = 1$
4) $(7 + 4\sqrt{3})(2 - \sqrt{3})^2 = (7 + 4\sqrt{3})(2^2 - 2 * 2\sqrt{3} + (\sqrt{3})^2) = (7 + 4\sqrt{3})(4 - 4\sqrt{3} + 3) = (7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - 16 * 3 = 49 - 48 = 1$
5) $(\sqrt{6 + 2\sqrt{5}} - \sqrt{6 - 2\sqrt{5}})^2 = (\sqrt{6 + 2\sqrt{5}})^2 - 2\sqrt{6 + 2\sqrt{5}}\sqrt{6 - 2\sqrt{5}} + (\sqrt{6 - 2\sqrt{5}})^2 = 6 + 2\sqrt{5} - 2\sqrt{(6 - 2\sqrt{5})(6 + 2\sqrt{5})} + 6 - 2\sqrt{5} = 12 - 2\sqrt{6^2 - (2\sqrt{5})^2} = 12 - 2\sqrt{36 - 4 * 5} = 12 - 2\sqrt{36 - 20} = 12 - 2\sqrt{16} = 12 - 2 * 4 = 12 - 8 = 4$
554. Найдите значение выражения:
1) $(3\sqrt{2} + 1)(\sqrt{8} - 2)$;
2) $(3 - 2\sqrt{7})^2 + (3 + 2\sqrt{7})^2$;
3) $(10 - 4\sqrt{6})(2 + \sqrt{6})^2$;
4) $(\sqrt{9 - 4\sqrt{2}} + \sqrt{9 + 4\sqrt{2}})^2$.
Решение:
1) $(3\sqrt{2} + 1)(\sqrt{8} - 2) = (3\sqrt{2} + 1)(\sqrt{4 * 2} - 2) = (3\sqrt{2} + 1)(2\sqrt{2} - 2) = 3\sqrt{2} * 2\sqrt{2} - 3\sqrt{2} * 2 + 1 * 2\sqrt{2} - 1 * 2 = 6 * 2 - 6\sqrt{2} + 2\sqrt{2} - 2 = 12 - 4\sqrt{2} - 2 = 10 - 4\sqrt{2}$
2) $(3 - 2\sqrt{7})^2 + (3 + 2\sqrt{7})^2 = 3^2 - 2 * 3 * 2\sqrt{7} + (2\sqrt{7})^2 + 3^2 + 2 * 3 * 2\sqrt{7} + (2\sqrt{7})^2 = 9 - 12\sqrt{7} + 4 * 7 + 9 + 12\sqrt{7} + 4 * 7 = 18 + 28 + 28 = 74$
3) $(10 - 4\sqrt{6})(2 + \sqrt{6})^2 = (10 - 4\sqrt{6})(2^2 + 2 * 2\sqrt{6} + (\sqrt{6})^2) = (10 - 4\sqrt{6})(4 + 4\sqrt{6} + 6) = (10 - 4\sqrt{6})(10 + 4\sqrt{6}) = 10^2 - (4\sqrt{6})^2 = 100 - 16 * 6 = 100 - 96 = 4$
4) $(\sqrt{9 - 4\sqrt{2}} + \sqrt{9 + 4\sqrt{2}})^2 = (\sqrt{9 - 4\sqrt{2}})^2 + 2\sqrt{9 - 4\sqrt{2}}\sqrt{9 + 4\sqrt{2}} + (\sqrt{9 + 4\sqrt{2}})^2 = 9 - 4\sqrt{2} + 2\sqrt{(9 - 4\sqrt{2})(9 + 4\sqrt{2})} + 9 + 4\sqrt{2} = 18 + 2\sqrt{9^2 - (4\sqrt{2})^2} = 18 + 2\sqrt{81 - 16 * 2} = 18 + 2\sqrt{81 - 32} = 18 + 2\sqrt{49} = 18 + 2 * 7 = 18 + 14 = 32$
555. Сократите дробь:
1) $\frac{4a + 4\sqrt{5}}{a^2 - 5}$;
2) $\frac{\sqrt{28} - 2\sqrt{2a}}{6a - 21}$;
3) $\frac{a + 4\sqrt{ab} + 4b}{a - 4b}$, если a > 0, b > 0;
4) $\frac{x^2 - 6y}{x^2 + 6y - x\sqrt{24y}}$;
5) $\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a^3} + \sqrt{b^3}}$;
6) $\frac{m\sqrt{m} - 27}{\sqrt{m} - 3}$.
Решение:
1) $\frac{4a + 4\sqrt{5}}{a^2 - 5} = \frac{4(a + \sqrt{5})}{(a - \sqrt{5})(a + \sqrt{5})} = \frac{4}{a -\sqrt{5}}$
2) $\frac{\sqrt{28} - 2\sqrt{2a}}{6a - 21} = \frac{\sqrt{4 * 7} - 2\sqrt{2a}}{3(2a - 7)} = \frac{2\sqrt{7} - 2\sqrt{2a}}{3((\sqrt{2a})^2 - (\sqrt{7})^2)} = \frac{2(\sqrt{7} - \sqrt{2a})}{3(\sqrt{2a} - \sqrt{7})(\sqrt{2a} + \sqrt{7})} = -\frac{2(\sqrt{2a} - \sqrt{7})}{3(\sqrt{2a} - \sqrt{7})(\sqrt{2a} + \sqrt{7})} = -\frac{2}{3(\sqrt{2a} + \sqrt{7})}$
3) $\frac{a + 4\sqrt{ab} + 4b}{a - 4b} = \frac{(\sqrt{a})^2 + 2 * \sqrt{a} * 2\sqrt{b} + (2\sqrt{b})^2}{(\sqrt{a})^2 - (2\sqrt{b})^2} = \frac{(\sqrt{a} + 2\sqrt{b})^2}{(\sqrt{a} - 2\sqrt{b})(\sqrt{a} + 2\sqrt{b})} = \frac{\sqrt{a} + 2\sqrt{b}}{\sqrt{a} - 2\sqrt{b}}$, если a > 0, b > 0
4) $\frac{x^2 - 6y}{x^2 + 6y - x\sqrt{24y}} = \frac{x^2 - (\sqrt{6y})^2}{x^2 - x\sqrt{4 * 6y} + (\sqrt{6y})^2} = \frac{(x - \sqrt{6y})(x + \sqrt{6y})}{x^2 - 2x\sqrt{6y} + (\sqrt{6y})^2} = \frac{(x - \sqrt{6y})(x + \sqrt{6y})}{(x - \sqrt{6y})^2} = \frac{x + \sqrt{6y}}{x - \sqrt{6y}}$
5) $\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a^3} + \sqrt{b^3}} = \frac{\sqrt{a} + \sqrt{b}}{(\sqrt{a})^3 + (\sqrt{b})^3} = \frac{\sqrt{a} + \sqrt{b}}{(\sqrt{a} + \sqrt{b})((\sqrt{a})^2 - \sqrt{a}\sqrt{b} + (\sqrt{b})^2)} = \frac{\sqrt{a} + \sqrt{b}}{(\sqrt{a} + \sqrt{b})(a - \sqrt{a}\sqrt{b} + b)} = \frac{1}{a - \sqrt{a}\sqrt{b} + b}$
6) $\frac{m\sqrt{m} - 27}{\sqrt{m} - 3} = \frac{\sqrt{m^2 * m} - 27}{\sqrt{m} - 3} = \frac{\sqrt{m^3} - 27}{\sqrt{m} - 3} = \frac{(\sqrt{m})^3 - 3^3}{\sqrt{m} - 3} = \frac{(\sqrt{m} - 3)((\sqrt{m})^2) + 3\sqrt{m} + 3^2)}{\sqrt{m} - 3} = m + 3\sqrt{m} + 9$
556. Сократите дробь:
1) $\frac{a - b}{\sqrt{11b} - \sqrt{11a}}$;
2) $\frac{2a + 10\sqrt{2ab} + 25b}{6a - 75b}$, если a > 0, b > 0;
3) $\frac{a - 2\sqrt{a} + 4}{a\sqrt{a} + 8}$.
Решение:
1) $\frac{a - b}{\sqrt{11b} - \sqrt{11a}} = \frac{(\sqrt{a})^2 - (\sqrt{b})^2}{\sqrt{11} * \sqrt{b} - \sqrt{11} * \sqrt{a}} = \frac{(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b})}{\sqrt{11}(\sqrt{b} - \sqrt{a})} = -\frac{(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b})}{\sqrt{11}(\sqrt{a} - \sqrt{b})} = -\frac{\sqrt{a} + \sqrt{b}}{\sqrt{11}}$
2) $\frac{2a + 10\sqrt{2ab} + 25b}{6a - 75b} = \frac{(\sqrt{2a})^2 + 10\sqrt{2ab} + (\sqrt{25b})^2}{3(2a - 25b)} = \frac{(\sqrt{2a})^2 + 2 * \sqrt{2a} * 5\sqrt{b} + (5\sqrt{b})^2}{3((\sqrt{2a})^2 - (\sqrt{25b})^2)} = \frac{(\sqrt{2a} + 5\sqrt{b})^2}{3((\sqrt{2a})^2 - (5\sqrt{b})^2)} = \frac{(\sqrt{2a} + 5\sqrt{b})^2}{3(\sqrt{2a} - 5\sqrt{b})(\sqrt{2a} + 5\sqrt{b})} = \frac{\sqrt{2a} + 5\sqrt{b}}{3(\sqrt{2a} - 5\sqrt{b})}$, если a > 0, b > 0
3) $\frac{a - 2\sqrt{a} + 4}{a\sqrt{a} + 8} = \frac{a - 2\sqrt{a} + 4}{\sqrt{a^2 * a} + 8} = \frac{a - 2\sqrt{a} + 4}{\sqrt{a^3} + 8} = \frac{a - 2\sqrt{a} + 4}{(\sqrt{a})^3 + 2^3} = \frac{a - 2\sqrt{a} + 4}{(\sqrt{a} + 2)((\sqrt{a})^2 - 2\sqrt{a} + 2^2)} = \frac{a - 2\sqrt{a} + 4}{(\sqrt{a} + 2)(a - 2\sqrt{a} + 4)} = \frac{1}{\sqrt{a} + 2}$
