Question

Rút gọn : K=\(\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\) với \(x>0,x\ne1\)

giúppp

Answer 1

Với \(x>0;x\ne1\) ta có:

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

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

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

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

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

\(=\sqrt{x}-1\)

Vậy \(K=\sqrt{x}-1.\)