Question

Rút gọn

a. \(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+x}{x-\sqrt{x}+1}+x+1\) với x > 0

b. \(\left(\dfrac{2+\sqrt{x}}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(\dfrac{x\sqrt{x}+x-\sqrt{x}-1}{\sqrt{x}}\right)\) với x > 0

Answer 1

a: Sửa đề: \(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\)

\(=x-\sqrt{x}-x-\sqrt{x}+x+1=x-2\sqrt{x}+1\)

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

\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}}=2\)