Question

Rút gọn

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

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

C=\(\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right)\dfrac{x+2\sqrt{x}}{\sqrt{x}}\)

Answer 1

a: \(A=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)

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

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