Question

Cho biểu thức \(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2}{x-4}\right).\left(\sqrt{x}-1+\dfrac{\sqrt{x}-4}{\sqrt{x}}\right)\left(với:x>0,x\ne4\right)\)
Chứng minh rằng \(P=\sqrt{x}+3\)

Answer 1

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

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

\(=\dfrac{x+3\sqrt{x}+2-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-\sqrt{x}+\sqrt{x}-4}{\sqrt{x}}\)

\(=\dfrac{x+3\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}}=\sqrt{x}+3\)