Question

rút gọn biểu thức :\(\frac{a-b}{b^2}\sqrt{\frac{a^2b^4}{a^2-2ab+b^2}}\)

 

Answer 1

Đặt A=\(\frac{a-b}{b^2}\sqrt{\frac{a^2b^4}{a^2-2ab+b^2}}=\frac{a-b}{b^2}\sqrt{\frac{a^2b^4}{\left(a-b\right)^2}}=\frac{a-b}{b^2}.\left|\frac{ab^2}{a-b}\right|\)

Với a<b thì : A=\(\frac{a-b}{b^2}.\frac{ab^2}{-\left(a-b\right)}=-a\)

Với a>b thì : A=\(\frac{a-b}{b^2}.\frac{ab^2}{a-b}=a\)

Answer 2

\(\frac{a-b}{b^2}\sqrt{\frac{a^2b^4}{a^2-2ab+b^2}}\)

\(=\frac{a-b}{b^2}\sqrt{\frac{\left(ab^2\right)^2}{\left(a-b\right)^2}}\)

\(=\frac{a-b}{b^2}\cdot\frac{\sqrt{\left(ab^2\right)^2}}{\sqrt{\left(a-b\right)^2}}\)

\(=\frac{a-b}{b^2}\cdot\frac{\left|a\right|b^2}{\left|a-b\right|}\)

+) Nếu a>b => \(\frac{a-b}{b^2}\cdot\frac{ab^2}{a-b}=a\)

+) Nếu a<b => \(\frac{a-b}{b^2}\cdot\frac{ab^2}{b-a}=-a\)