Tính: (x^2-2xy/y)*(y^2/x)

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

Bài 3 trang 25 SBT Toán 8 Tập 1:Tính:

a)$\frac{x^2-2xy}{y}.\frac{y^2}{x}$;

b)$\frac{x^2-9y^2}{3x{y}^2}.\frac{xy}{x+3y}$;

c)$\frac{1-x^2}{2x+4y}.\frac{x^2+4xy+4y^2}{3-3x}$;

d)$\frac{x^3-y^3}{x+y}.\frac{x^2-y^2}{x^2+xy+y^2}$.

Lời giải:

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

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

b)$\frac{x^2-9y^2}{3x{y}^2}.\frac{xy}{x+3y}=\frac{\left(x^2-9y^2\right).xy}{3x{y}^2.\left(x+3y\right)}$

$=\frac{\left(x+3y\right)\left(x-3y\right).xy}{3xy.y.\left(x+3y\right)}=\frac{x-3y}{3y}$.

c)$\frac{1-x^2}{2x+4y}.\frac{x^2+4xy+4y^2}{3-3x}$

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

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

$=\frac{\left(1+x\right)(x+2y)}{6}$.

d)$\frac{x^3-y^3}{x+y}.\frac{x^2-y^2}{x^2+xy+y^2}$

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

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