Giải hộ mình câu này với các bạn

Nguyễn Mai Phương

Bài 1:

$g,~\frac{-6x^2}{10y^3} \cdot \frac{5y}{9x^4} = \frac{-6x^2 \cdot 5y}{10y^3 \cdot 9x^4} = \frac{-30x^2y}{90x^4y^3} = -\frac{1}{3x^2y^2}$

$a,~\frac{x-1}{x+2} + \frac{3-x}{x+2} = \frac{x-1+3-x}{x+2} = \frac{2}{x+2}$

$b,~\frac{2y}{x-y} + \frac{2x}{y-x} = \frac{2y}{x-y} - \frac{2x}{x-y} = \frac{2y-2x}{x-y} = \frac{-2(x-y)}{x-y} = -2$

$h,~\frac{x-5}{x} \cdot \frac{5x}{x^2-25} = \frac{x-5}{x} \cdot \frac{5x}{(x-5)(x+5)} = \frac{5}{x+5}$

$c,~\frac{12}{x^2-9} - \frac{2}{x-3} = \frac{12}{(x-3)(x+3)} - \frac{2(x+3)}{(x-3)(x+3)} = \frac{12 - 2x - 6}{(x-3)(x+3)} = \frac{6-2x}{(x-3)(x+3)} = \frac{-2(x-3)}{(x-3)(x+3)} = \frac{-2}{x+3}$

$i,~\frac{-6x}{4y^3} : \left(-\frac{x^4}{24y}\right) = \frac{-6x}{4y^3} \cdot \left(-\frac{24y}{x^4}\right) = \frac{144xy}{4x^4y^3} = \frac{36}{x^3y^2}$

$d,~\frac{6x+2}{4x-4} + \frac{3-9x}{6x-6} = \frac{6x+2}{4(x-1)} + \frac{3-9x}{6(x-1)} = \frac{3(6x+2)}{12(x-1)} + \frac{2(3-9x)}{12(x-1)} = \frac{18x+6 + 6-18x}{12(x-1)} = \frac{12}{12(x-1)} = \frac{1}{x-1}$

$k,~\frac{x^2-9}{x+3} : \frac{2x-6}{x} = \frac{(x-3)(x+3)}{x+3} \cdot \frac{x}{2(x-3)} = (x-3) \cdot \frac{x}{2(x-3)} = \frac{x}{2}$

$e,~\frac{2}{xy-x^2} - \frac{2}{y^2-xy} = \frac{2}{x(y-x)} - \frac{2}{y(y-x)} = \frac{2y}{xy(y-x)} - \frac{2x}{xy(y-x)} = \frac{2y-2x}{xy(y-x)} = \frac{2(y-x)}{xy(y-x)} = \frac{2}{xy}$

$B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right) \cdot \frac{4x^2-4}{5}$

Bài 2:

a, ĐKXĐ: $x \neq 1$ và $x \neq -1$.

$b, B=\left(\frac{x+1}{2(x-1)} + \frac{3}{(x-1)(x+1)} - \frac{x+3}{2(x+1)}\right) \cdot \frac{4(x^2-1)}{5} $

$B=\left(\frac{(x+1)^2}{2(x-1)(x+1)} + \frac{3 \cdot 2}{2(x-1)(x+1)} - \frac{(x+3)(x-1)}{2(x-1)(x+1)}\right) \cdot \frac{4(x-1)(x+1)}{5} $

$B=\left(\frac{x^2+2x+1 + 6 - (x^2 + 2x - 3)}{2(x^2-1)}\right) \cdot \frac{4(x^2-1)}{5}$

$B=\left(\frac{x^2+2x+7 - x^2 - 2x + 3}{2(x^2-1)}\right) \cdot \frac{4(x^2-1)}{5}$

$B=\frac{10}{2(x^2-1)} \cdot \frac{4(x^2-1)}{5}$

$B=\frac{5}{x^2-1} \cdot \frac{4(x^2-1)}{5} = 4$