`a)`\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(ĐK:\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(A=\left(\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{2}\)
\(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(A=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(A=\dfrac{2}{x+\sqrt{x}+1}\)
`b)`\(x=7-2\sqrt{6}=\left(\sqrt{6}-1\right)^2\)
\(\Rightarrow A=\dfrac{2}{7-2\sqrt{6}+\sqrt{\left(\sqrt{6}-1\right)^2}+1}\)
\(A=\dfrac{2}{7-2\sqrt{6}+\sqrt{6}-1+1}\)
\(A=\dfrac{2}{7-\sqrt{6}}\)
a, đk x >= 0 ; x khác 1
\(A=\left(\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{x\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x-2\sqrt{x}+1}{x\sqrt{x}-1}:\dfrac{\sqrt{x}-1}{2}=\dfrac{2}{x+\sqrt{x}+1}\)
b, Ta có \(\sqrt{x}=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)
\(\dfrac{2}{7-2\sqrt{6}+\sqrt{6}-1}=\dfrac{2}{6-\sqrt{6}}=\dfrac{2\left(6+\sqrt{6}\right)}{30}=\dfrac{6+\sqrt{6}}{15}\)
