Question

Chứng minh:

\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2=1\) với \(a\ge0,a\ne1\)

Answer 1

\(VT=\dfrac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{1-\sqrt{x}}.\left(\dfrac{1-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)^2\)

\(=\left(1+\sqrt{x}+x\right).\left(\dfrac{1}{1+\sqrt{x}}\right)^2\)

\(=\left(1+\sqrt{x}+x\right).\dfrac{1}{\left(1+\sqrt{x}\right)^2}\)

\(=\dfrac{1+\sqrt{x}+x}{\left(1+\sqrt{x}\right)^2}=\dfrac{\left(1+\sqrt{x}\right)^2}{\left(1+\sqrt{x}\right)^2}=1=VP\)