Giải bài 2.32 trang 52

Đề bài

Thực hiện các phép tính sau:

a)     $\frac{y^2-4y+4}{3-9y}.\frac{3y-1}{3y^2-12}$

b)    $\frac{c^2-d^2}{cd}:\frac{1}{cd+d^2}$

Lời giải chi tiết

a)    

$\begin{array}{c}\frac{y^2-4y+4}{3-9y}.\frac{3y-1}{3y^2-12}\\ =\frac{\left(y^2-4y+4\right)\left(3y-1\right)}{\left(3-9y\right)\left(3y^2-12\right)}\\ =\frac{{\left(y-2\right)}^2\left(3y-1\right)}{-3\left(3y-1\right).3\left(y-2\right)\left(y+2\right)}\\ =\frac{\left(y-2\right)}{-9\left(y+2\right)}\end{array}$

b)   

$\begin{array}{c}\frac{c^2-d^2}{cd}:\frac{1}{cd+d^2}\\ =\frac{c^2-d^2}{cd}.\frac{cd+d^2}{1}\\ =\frac{\left(c^2-d^2\right).\left(cd+d^2\right)}{cd}\\ =\frac{\left(c^2-d^2\right).\left(c+d\right)d}{cd}\\ =\frac{\left(c^2-d^2\right).\left(c+d\right)}{c}\end{array}$