Question

RÚT GỌN BT :

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

Answer 1

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