Question

[Lưu Thị Tuyết]

  

Answer 1

[Nguyễn Việt Lâm]

ĐKXĐ: \(x\ge0;x\ne1\)

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

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

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

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

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

Answer 2

[Nguyễn Tân Vương]

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

\(ĐKXĐ:x\ge0.x\ne1\)

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

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

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

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

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

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

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

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

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