Question

rút gọn biểu thức sau:

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

Answer 1

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

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

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