Ответы к странице 30

120. Докажите тождество:
1) $\frac{1}{6a - 4b} - \frac{1}{6a + 4b} - \frac{3a}{4b^2 - 9a^2} = \frac{1}{3a - 2b}$;
2) $\frac{c + 2}{c^2 + 3c} - \frac{1}{3c + 9} - \frac{2}{3c} = 0$.

Решение:

1) $\frac{1}{6a - 4b} - \frac{1}{6a + 4b} - \frac{3a}{4b^2 - 9a^2} = \frac{1}{3a - 2b}$
$\frac{1}{6a - 4b} - \frac{1}{6a + 4b} + \frac{3a}{9a^2 - 4b^2} = \frac{1}{3a - 2b}$
$\frac{1}{2(3a - 2b)} - \frac{1}{2(3a + 2b)} + \frac{3a}{(3a - 2b)(3a + 2b)} = \frac{1}{3a - 2b}$
$\frac{3a + 2b - (3a - 2b) + 2 * 3a}{2(3a - 2b)(3a + 2b)} = \frac{1}{3a - 2b}$
$\frac{3a + 2b - 3a + 2b + 6a}{2(3a - 2b)(3a + 2b)} = \frac{1}{3a - 2b}$
$\frac{6a + 4b}{2(3a - 2b)(3a + 2b)} = \frac{1}{3a - 2b}$
$\frac{2(3a + 2b)}{2(3a - 2b)(3a + 2b)} = \frac{1}{3a - 2b}$
$\frac{1}{3a - 2b} = \frac{1}{3a - 2b}$

2) $\frac{c + 2}{c^2 + 3c} - \frac{1}{3c + 9} - \frac{2}{3c} = 0$
$\frac{c + 2}{c(c + 3)} - \frac{1}{3(c + 3)} - \frac{2}{3c} = 0$
$\frac{3(c + 2) - c - 2(c + 3)}{3c(c + 3)} = 0$
$\frac{3c + 6 - c - 2c - 6}{3c(c + 3)} = 0$
$\frac{0}{3c(c + 3)} = 0$
0 = 0

121. Найдите разность дробей:
1) $\frac{a + 1}{a^3 - 1} - \frac{1}{a^2 + a + 1}$;
2) $\frac{1}{b + 3} - \frac{b^2 - 6b}{b^3 + 27}$.

Решение:

1) $\frac{a + 1}{a^3 - 1} - \frac{1}{a^2 + a + 1} = \frac{a + 1}{(a - 1)(a^2 + a + 1)} - \frac{1}{a^2 + a + 1} = \frac{a + 1 - (a - 1)}{(a - 1)(a^2 + a + 1)} = \frac{a + 1 - a + 1}{(a - 1)(a^2 + a + 1)} = \frac{2}{a^3 - 1}$

2) $\frac{1}{b + 3} - \frac{b^2 - 6b}{b^3 + 27} = \frac{1}{b + 3} - \frac{b^2 - 6b}{(b + 3)(b^2 - 3b + 9)} = \frac{b^2 - 3b + 9 - (b^2 - 6b)}{(b + 3)(b^2 - 3b + 9)} = \frac{b^2 - 3b + 9 - b^2 + 6b}{(b + 3)(b^2 - 3b + 9)} = \frac{3b + 9}{(b + 3)(b^2 - 3b + 9)} = \frac{3(b + 3)}{(b + 3)(b^2 - 3b + 9)} = \frac{3}{b^2 - 3b + 9}$

122. Упростите выражение:
1) $\frac{9m^2 - 3mn + n^2}{3m - n} - \frac{9m^2 + 3mn + n^2}{3m + n}$;
2) $1 - \frac{2b - 1}{4b^2 - 2b + 1} - \frac{2b}{2b + 1}$.

Решение:

1) $\frac{9m^2 - 3mn + n^2}{3m - n} - \frac{9m^2 + 3mn + n^2}{3m + n} = \frac{(3m + n)(9m^2 - 3mn + n^2) - (3m - n)(9m^2 + 3mn + n^2)}{(3m - n)(3m + n)} = \frac{81m^3 + n^3 - (81m^3 - n^3)}{(3m - n)(3m + n)} = \frac{81m^3 + n^3 - 81m^3 + n^3}{(3m - n)(3m + n)} = \frac{2n^3}{9m^2 - n^2}$

2) $1 - \frac{2b - 1}{4b^2 - 2b + 1} - \frac{2b}{2b + 1} = \frac{(2b + 1)(4b^2 - 2b + 1) - (2b - 1)(2b + 1) - 2b(4b^2 - 2b + 1)}{(2b + 1)(4b^2 - 2b + 1)} = \frac{8b^3 + 1 - (4b^2 - 1) - 8b^3 + 4b^2 - 2b}{(2b + 1)(4b^2 - 2b + 1)} = \frac{8b^3 + 1 - 4b^2 + 1 - 8b^3 + 4b^2 - 2b}{(2b + 1)(4b^2 - 2b + 1)} = \frac{2 - 2b}{8b^3 + 1}$

123. Докажите тождество:
$\frac{3a^2 + 24}{a^3 + 8} - \frac{6}{a^2 - 2a + 4} - \frac{1}{a + 2} = \frac{2}{a + 2}$.

Решение:

$\frac{3a^2 + 24}{a^3 + 8} - \frac{6}{a^2 - 2a + 4} - \frac{1}{a + 2} = \frac{2}{a + 2}$
$\frac{3a^2 + 24}{(a + 2)(a^2 - 2a + 4)} - \frac{6}{a^2 - 2a + 4} - \frac{1}{a + 2} = \frac{2}{a + 2}$
$\frac{3a^2 + 24 - 6(a + 2) - (a^2 - 2a + 4)}{(a + 2)(a^2 - 2a + 4)} = \frac{2}{a + 2}$
$\frac{3a^2 + 24 - 6a - 12 - a^2 + 2a - 4}{(a + 2)(a^2 - 2a + 4)} = \frac{2}{a + 2}$
$\frac{2a^2 - 4a + 8}{(a + 2)(a^2 - 2a + 4)} = \frac{2}{a + 2}$
$\frac{2(a^2 - 2a + 4)}{(a + 2)(a^2 - 2a + 4)} = \frac{2}{a + 2}$
$\frac{2}{a + 2} = \frac{2}{a + 2}$

124. Упростите выражение:
1) $\frac{4b}{a^2 - b^2} + \frac{a - b}{a^2 + ab} + \frac{a + b}{b^2 - ab}$;
2) $\frac{1}{x - 2} + \frac{1}{x + 2} - \frac{x}{x^2 - 4} + \frac{x^2 + 4}{8x - 2x^3}$;
3) $\frac{1}{(a - 5b)^2} - \frac{2}{a^2 - 25b^2} + \frac{1}{(a + 5b)^2}$;
4) $\frac{x^2 + 9x + 18}{xy + 3y - 2x - 6} - \frac{x + 5}{y - 2}$.

Решение:

1) $\frac{4b}{a^2 - b^2} + \frac{a - b}{a^2 + ab} + \frac{a + b}{b^2 - ab} = \frac{4b}{a^2 - b^2} + \frac{a - b}{a^2 + ab} - \frac{a + b}{ab - b^2} = \frac{4b}{(a - b)(a + b)} + \frac{a - b}{a(a + b)} - \frac{a + b}{b(a - b)} = \frac{4b * ab + b(a - b)(a - b) - a(a + b)(a + b)}{ab(a - b)(a + b)} = \frac{4ab^2 + b(a - b)^2 - a(a + b)^2}{ab(a - b)(a + b)} = \frac{4ab^2 + b(a^2 - 2ab + b^2) - a(a^2 + 2ab + b^2)}{ab(a - b)(a + b)} = \frac{4ab^2 + a^2b - 2ab^2 + b^3 - a^3 - 2a^2b - ab^2}{ab(a - b)(a + b)} = \frac{ab^2 - a^2b + b^3 - a^3}{ab(a - b)(a + b)} = \frac{(ab^2 - a^2b) + (b^3 - a^3)}{ab(a - b)(a + b)} = \frac{ab(b - a) + (b - a)(b^2 + ab + a^2)}{ab(a - b)(a + b)} = \frac{(b - a)(ab + b^2 + ab + a^2}{ab(a - b)(a + b)} = \frac{-(a - b)(a^2 + 2ab + b^2)}{ab(a - b)(a + b)} = \frac{-(a - b)(a + b)^2}{ab(a - b)(a + b)} = \frac{-(a + b)}{ab} = -\frac{a + b}{ab}$

2) $\frac{1}{x - 2} + \frac{1}{x + 2} - \frac{x}{x^2 - 4} + \frac{x^2 + 4}{8x - 2x^3} = \frac{1}{x - 2} + \frac{1}{x + 2} - \frac{x}{x^2 - 4} - \frac{x^2 + 4}{2x^3 - 8x} = \frac{1}{x - 2} + \frac{1}{x + 2} - \frac{x}{(x - 2)(x + 2)} - \frac{x^2 + 4}{2x(x^2 - 4)} = \frac{1}{x - 2} + \frac{1}{x + 2} - \frac{x}{(x - 2)(x + 2)} - \frac{x^2 + 4}{2x(x - 2)(x + 2)} = \frac{2x(x + 2) + 2x(x - 2) - 2x * x - (x^2 + 4)}{2x(x - 2)(x + 2)} = \frac{2x^2 + 4x + 2x^2 - 4x - 2x^2 - x^2 - 4}{2x(x - 2)(x + 2)} = \frac{x^2 - 4}{2x(x^2 - 4)} = \frac{1}{2x}$

3) $\frac{1}{(a - 5b)^2} - \frac{2}{a^2 - 25b^2} + \frac{1}{(a + 5b)^2} = \frac{1}{(a - 5b)^2} - \frac{2}{(a - 5b)(a + 5b)} + \frac{1}{(a + 5b)^2} = \frac{(a + 5b)^2 - 2(a^2 - 25b^2) + (a - 5b)^2}{(a - 5b)^2(a + 5b)^2} = \frac{a^2 + 10ab + 25b^2 - 2a^2 + 50b^2 + a^2 - 10ab + 25b^2}{(a - 5b)^2(a + 5b)^2} = \frac{100b^2}{(a^2 - 25b^2)^2}$

4) $\frac{x^2 + 9x + 18}{xy + 3y - 2x - 6} - \frac{x + 5}{y - 2} = \frac{x^2 + 9x + 18}{(xy + 3y) - (2x + 6)} - \frac{x + 5}{y - 2} = \frac{x^2 + 9x + 18}{y(x + 3) - 2(x + 3)} - \frac{x + 5}{y - 2} = \frac{x^2 + 9x + 18}{(x + 3)(y - 2)} - \frac{x + 5}{y - 2} = \frac{x^2 + 9x + 18 - (x + 3)(x + 5)}{(x + 3)(y - 2)} = \frac{x^2 + 9x + 18 - (x^2 + 3x + 5x + 15)}{(x + 3)(y - 2)} = \frac{x^2 + 9x + 18 - x^2 - 3x - 5x - 15}{(x + 3)(y - 2)} = \frac{x + 3}{(x + 3)(y - 2)} = \frac{1}{y - 2}$

125. Докажите тождество:
1) $\frac{a + 3}{a^2 - 3a} + \frac{a - 3}{3a + 9} + \frac{12}{9 - a^2} = \frac{a - 3}{3a}$;
2) $\frac{b - 4}{2a - 1} - \frac{b^2 - 2b - 24}{2ab - 4 - b + 8a} = \frac{2}{2a - 1}$.

Решение:

1) $\frac{a + 3}{a^2 - 3a} + \frac{a - 3}{3a + 9} + \frac{12}{9 - a^2} = \frac{a - 3}{3a}$
$\frac{a + 3}{a^2 - 3a} + \frac{a - 3}{3a + 9} - \frac{12}{a^2 - 9} = \frac{a - 3}{3a}$
$\frac{a + 3}{a(a - 3)} + \frac{a - 3}{3(a + 3)} - \frac{12}{(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{3(a + 3)^2 + a(a - 3)^2 - 3a * 12}{3a(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{3(a^2 + 6a + 9) + a(a^2 - 6a + 9) - 36a}{3a(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{3a^2 + 18a + 27 + a^3 - 6a^2 + 9a - 36a}{3a(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{a^3 - 3a^2 - 9a + 27}{3a(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{(a^3 - 3a^2) - (9a - 27)}{3a(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{a^2(a - 3) - 9(a - 3)}{3a(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{(a - 3)(a^2 - 9)}{3a(a^2 - 9)} = \frac{a - 3}{3a}$
$\frac{a - 3}{3a} = \frac{a - 3}{3a}$

2) $\frac{b - 4}{2a - 1} - \frac{b^2 - 2b - 24}{2ab - 4 - b + 8a} = \frac{2}{2a - 1}$
$\frac{b - 4}{2a - 1} - \frac{b^2 - 2b - 24}{(2ab - b) + (8a - 4)} = \frac{2}{2a - 1}$
$\frac{b - 4}{2a - 1} - \frac{b^2 - 2b - 24}{b(2a - 1) + 4(2a - 1)} = \frac{2}{2a - 1}$
$\frac{b - 4}{2a - 1} - \frac{b^2 - 2b - 24}{(2a - 1)(b + 4)} = \frac{2}{2a - 1}$
$\frac{(b - 4)(b + 4) - (b^2 - 2b - 24)}{(2a - 1)(b + 4)} = \frac{2}{2a - 1}$
$\frac{b^2 - 4b + 4b - 16 - b^2 + 2b + 24}{(2a - 1)(b + 4)} = \frac{2}{2a - 1}$
$\frac{2b + 8}{(2a - 1)(b + 4)} = \frac{2}{2a - 1}$
$\frac{2(b + 4)}{(2a - 1)(b + 4)} = \frac{2}{2a - 1}$
$\frac{2}{2a - 1} = \frac{2}{2a - 1}$

126. Докажите тождество:
$\frac{1}{(a - b)(a - c)} - \frac{1}{(a - b)(b - c)} + \frac{1}{(c - a)(c - b)} = 0$

Решение:

$\frac{1}{(a - b)(a - c)} - \frac{1}{(a - b)(b - c)} + \frac{1}{(c - a)(c - b)} = 0$
$\frac{1}{(a - b)(a - c)} - \frac{1}{(a - b)(b - c)} + \frac{1}{(a - c)(b - c)} = 0$
$\frac{b - c - (a - c) + a - b}{(a - b)(a - c)(b - c)} = 0$
$\frac{b - c - a + c + a - b}{(a - b)(a - c)(b - c)} = 0$
$\frac{0}{(a - b)(a - c)(b - c)} = 0$
0 = 0

127. Докажите тождество:
$\frac{bc}{(a - b)(a - c)} + \frac{ac}{(b - a)(b - c)} + \frac{ab}{(c - a)(c - b)} = 1$

Решение:

$\frac{bc}{(a - b)(a - c)} + \frac{ac}{(b - a)(b - c)} + \frac{ab}{(c - a)(c - b)} = 1$
$\frac{bc}{(a - b)(a - c)} - \frac{ac}{(a - b)(b - c)} + \frac{ab}{(a - c)(b - c)} = 1$
$\frac{bc(b - c) - ac(a - c) + ab(a - b)}{(a - b)(a - c)(b - c)} = 1$
$\frac{b^2c - bc^2 - a^2c + ac^2 + a^2b - ab^2}{(a - b)(a - c)(b - c)} = 1$
$\frac{b^2c - bc^2 - a^2c + ac^2 + a^2b - ab^2}{(a - b)(a - c)(b - c)} = 1$
$\frac{(b^2c - ab^2) - (bc^2 - a^2b) - (a^2c - ac^2)}{(a - b)(a - c)(b - c)} = 1$
$\frac{b^2(c - a) - b(c^2 - a^2) - ac(a - c)}{(a - b)(a - c)(b - c)} = 1$
$\frac{b^2(c - a) - b(c - a)(c + a) + ac(c - a)}{(a - b)(a - c)(b - c)} = 1$
$\frac{(c - a)(b^2 - b(c + a) + ac)}{(a - b)(a - c)(b - c)} = 1$
$\frac{(c - a)(b^2 - bc - ab + ac)}{(a - b)(a - c)(b - c)} = 1$
$\frac{(c - a)((b^2 - bc) - (ab - ac))}{(a - b)(a - c)(b - c)} = 1$
$\frac{(c - a)(b(b - c) - a(b - c))}{(a - b)(a - c)(b - c)} = 1$
$\frac{(c - a)(b - c)(b - a)}{(a - b)(a - c)(b - c)} = 1$
$\frac{(a - c)(b - c)(a - b)}{(a - b)(a - c)(b - c)} = 1$
1 = 1

128. Упростите выражение:
$\frac{1}{(a - 1)(a - 2)} + \frac{1}{(a - 2)(a - 3)} + \frac{1}{(a - 3)(a - 4)}$.

Решение:

$\frac{1}{(a - 1)(a - 2)} + \frac{1}{(a - 2)(a - 3)} + \frac{1}{(a - 3)(a - 4)} = \frac{(a - 3)(a - 4) + (a - 1)(a - 4) + (a - 1)(a - 2)}{(a - 1)(a - 2)(a - 3)(a - 4)} = \frac{a^2 - 3a - 4a + 12 + a^2 - a - 4a + 4 + a^2 - a - 2a + 2}{(a - 1)(a - 2)(a - 3)(a - 4)} = \frac{3a^2 - 15a + 18}{(a - 1)(a - 2)(a - 3)(a - 4)} = \frac{3(a^2 - 5a + 6)}{(a - 1)(a^2 - 2a - 3a + 6)(a - 4)} = \frac{3(a^2 - 5a + 6)}{(a - 1)(a^2 - 5a + 6)(a - 4)} = \frac{3}{(a - 1)(a - 4)}$