\(a,\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}+2}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\left(dkxd:x\ge0;x\ne1\right)\)
\(=\left[\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\left(\sqrt{x}-1\right)\)
\(=\dfrac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)\)
\(=\dfrac{\left(x+2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\)
\(=\sqrt{x}+1\)
\(b,\) Thay \(x=4-2\sqrt{3}\) vào biểu thức trên, ta được:
\(\sqrt{4-2\sqrt{3}}+1\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}+1\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+1\)
\(=\left|\sqrt{3}-1\right|+1\)
\(=\sqrt{3}-1+1\)
\(=\sqrt{3}\)
Vậy: ...
\(\text{#}Toru\)
\(a\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}+2}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\\ =\left(\dfrac{\sqrt{x}-1}{x-1}+\dfrac{x+\sqrt{x}+2}{x-1}\right).\sqrt{x}-1\\ =\dfrac{x+\sqrt{2}+1}{x-1}.\sqrt{x}-1\\ =\sqrt{x}+1\\ b,tacóx=4-2\sqrt{3}=\left(\sqrt{3}-\sqrt{1}\right)^2thãy=\sqrt{3}-\sqrt{1}vàobiểuthức,tađược\\ \sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}-1=\sqrt{3}-1-1=\sqrt{3}-2\)