Question

1. Cho biểu thức

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

a Rút gọn M

b. Tìm M khi x = \(6-2\sqrt{5}\)

Answer 1

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

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

\(=\dfrac{\left(1-x\right)+\sqrt{x}\left(1-x\right)}{\sqrt{x}-x}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1-x}{\sqrt{x}-x}=\dfrac{x-1}{x-\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

b: Khi x=6-2căn 5 thì \(M=\dfrac{\sqrt{5}-1+1}{\sqrt{5}-1}=\dfrac{\sqrt{5}}{\sqrt{5}-1}\)