a, ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\x\ge0\\\sqrt{x}+1\ne0\\x\sqrt{x}+x+\sqrt{x}+1\ne0\end{matrix}\right.\)
=> \(x\ge0\)
Ta có : \(B=\left(\frac{1}{\sqrt{x}+1}-\frac{2}{x\sqrt{x}+x+\sqrt{x}+1}\right):\left(2-\frac{2x-\sqrt{x}}{x+1}\right)\)
=> \(B=\left(\frac{x+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\frac{2}{\left(x+1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2\left(x+1\right)}{x+1}-\frac{\sqrt{x}\left(2\sqrt{x}-1\right)}{x+1}\right)\)
=> \(B=\left(\frac{x+1-2}{\left(\sqrt{x}+1\right)\left(x+1\right)}\right):\left(\frac{2\left(x+1\right)-\sqrt{x}\left(2\sqrt{x}-1\right)}{x+1}\right)\)
=> \(B=\left(\frac{x+1-2}{\left(\sqrt{x}+1\right)\left(x+1\right)}\right):\left(\frac{2x+2-2x+\sqrt{x}}{x+1}\right)\)
=> \(B=\left(\frac{x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\right):\left(\frac{2+\sqrt{x}}{x+1}\right)\)
=> \(B=\frac{\left(x-1\right)\left(x+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)\left(\sqrt{x}+2\right)}\)
=> \(B=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
- Ta có : \(x=\sqrt{6+2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)^2}\)
=> \(x=\sqrt{5}+1\)
- Thay\(x=\sqrt{5}+1\) vào phương trình ta được :
\(B=\frac{\sqrt{5}+1-1}{\sqrt{5}+1+2}=\frac{3\sqrt{5}-5}{4}\)
b, Ta có : \(B=\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\)
- Để B là số nguyên thì : \(\sqrt{x}+2\inƯ_{\left(3\right)}\)
=> x = 1 .
