Question

  

Answer 1

\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{1-\sqrt{x}}+\dfrac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\left(x\ge1\right)\)

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

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

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

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

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