Giải bài 2.33 trang 52

Đề bài

Rút gọn các biểu thức sau:

a)     $\left(b-\frac{a^2+b^2}{a+b}\right).\left(\frac{2b}{a}-\frac{4b}{a-b}\right)$

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

Lời giải chi tiết

a)    

$\begin{array}{c}\left(b-\frac{a^2+b^2}{a+b}\right).\left(\frac{2b}{a}-\frac{4b}{a-b}\right)\\ =\left(\frac{b\left(a+b\right)-a^2-b^2}{a+b}\right).\left(\frac{2b\left(a-b\right)-4ab}{a\left(a-b\right)}\right)\\ =\left(\frac{ab+b^2-a^2-b^2}{a+b}\right).\left(\frac{2ab-2b^2-4ab}{a\left(a-b\right)}\right)\\ =\left(\frac{ab-a^2}{a+b}\right).\left(\frac{2b^2-2ab}{a\left(a-b\right)}\right)\\ =\left(\frac{ab-a^2}{a+b}\right).\left(\frac{-2b\left(a-b\right)}{a\left(a-b\right)}\right)\\ =\left(\frac{ab-a^2}{a+b}\right).\frac{-2b}{a}\\ =\frac{\left(ab-a^2\right).-2b}{a\left(a+b\right)}\\ =\frac{-2a{b}^2+2a^{2}b}{a^2+ab}\\ =\frac{2ab-2b}{a+b}\end{array}$

b)   

$\begin{array}{c}\left(\frac{x^2}{y^2}+\frac{y}{x}\right):\left(\frac{x}{y^2}-\frac{1}{y}+\frac{1}{x}\right)\\ =\frac{x^3+y^3}{x{y}^2}:\frac{x^2-xy+y^2}{x{y}^2}\\ =\frac{x^3+y^3}{x{y}^2}.\frac{x{y}^2}{x^2-xy+y^2}\\ =x+y\end{array}$