Question

Answer 1

g: Ta có: \(G=\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\cdot\dfrac{2}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}-1+x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot2\)

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

h: Ta có: \(H=\left(\dfrac{3}{\sqrt{x}-2}+\dfrac{1}{x-4}\right):\dfrac{x-1}{\sqrt{x}+2}\)

\(=\dfrac{3\sqrt{x}+6+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{x-1}\)

\(=\dfrac{3\sqrt{x}+7}{\left(x-1\right)\left(\sqrt{x}-2\right)}\)