Question

Cho A = \(\dfrac{x+2}{x\sqrt{x}-1}\) + \(\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\) + \(\dfrac{1}{1-\sqrt{x}}\) với x \(\ge\) 0, x\(\ne\) 1

a, Rút gọn A

Answer 1

ĐKXĐ: \(x\ge0,x\ne1\)

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

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

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

\(A=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

Answer 2

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