Tính: (1/y+2/x-y).(x- x^2+y^2/x+y)

Giải SBT Toán 8 Bài 7: Nhân, chia phân thức

Bài 8 trang 25 SBT Toán 8 Tập 1: Tính:

a) $\left(\frac{1}{y}+\frac{2}{x-y}\right)\left(x-\frac{x^2+y^2}{x+y}\right)$;

b) $\left(\frac{x}{x+1}+1\right):\left(1-\frac{3x^2}{1-x^2}\right)$.

Lời giải:

a) $\left(\frac{1}{y}+\frac{2}{x-y}\right)\left(x-\frac{x^2+y^2}{x+y}\right)$

$=\left(\frac{x-y}{y(x-y)}+\frac{2y}{y\left(x-y\right)}\right)\left(\frac{x(x+y)}{x+y}-\frac{x^2+y^2}{x+y}\right)$

$=\frac{x-y+2y}{y(x-y)}.\frac{x^2+xy-x^2-y^2}{x+y}$

$=\frac{x+y}{y(x-y)}.\frac{xy-y^2}{x+y}=\frac{\left(x+y\right).y(x-y)}{y(x-y).\left(x+y\right)}=1$.

b) $\left(\frac{x}{x+1}+1\right):\left(1-\frac{3x^2}{1-x^2}\right)$

$=\left(\frac{x}{x+1}+\frac{x+1}{x+1}\right):\left(\frac{1-x^2}{1-x^2}-\frac{3x^2}{1-x^2}\right)$

$=\frac{x+x+1}{x+1}:\frac{1-x^2-3x^2}{1-x^2}$

$=\frac{2x+1}{x+1}:\frac{1-4x^2}{1-x^2}=\frac{2x+1}{x+1}.\frac{1-x^2}{1-4x^2}$

$=\frac{2x+1}{x+1}.\frac{\left(1+x\right)\left(1-x\right)}{\left(1+2x\right)(1-2x)}$

$=\frac{1-x}{1-2x}=\frac{-(x-1)}{-(2x-1)}=\frac{x-1}{2x-1}$.