Question

B= \(\dfrac{\sqrt{x}\left(1-x\right)^2}{1+x}\):\(\left[\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right).\left(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}-\sqrt{x}\right)\right]\)

a) Rút gọn B

b)Tìm x để B=\(\dfrac{2}{5}\)

Answer 1

a: \(B=\dfrac{\sqrt{x}\left(x-1\right)^2}{x+1}:\left[\left(x+\sqrt{x}+\sqrt{x}+1\right)\cdot\left(x-\sqrt{x}-\sqrt{x}+1\right)\right]\)

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

b: Để \(B=\dfrac{2}{5}\) thì \(\dfrac{\sqrt{x}}{x+1}=\dfrac{2}{5}\)

=>\(2x-5\sqrt{x}+2=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)

=>x=1/4 hoặc x=4