a/ đk: \(x>0;x\ne\)1
\(A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\frac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)\(:\left(\frac{\sqrt{x}+1}{\sqrt{x}+1}-\frac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left[\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\right]:\frac{2\sqrt{x}-2}{\sqrt{x}+1}\)\(=\frac{2\sqrt{x}}{\sqrt{x}}.\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
vậy A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\) với x>0;x\(\ne1\)
b/ ta có: \(x=\frac{1}{6-2\sqrt{5}}=\frac{1}{\left(\sqrt{5}-1\right)^2}\Rightarrow\sqrt{x}=\frac{1}{\sqrt{5}-1}\)
thay x=\(\frac{1}{\sqrt{5}-1}\left(TM\right)\) vào biểu thức A ta có:
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\frac{1}{\sqrt{5}-1}+1}{\frac{1}{\sqrt{5}-1}-1}=-5-2\sqrt{5}\)
vậy A=\(-5-2\sqrt{5}\) tại x=\(\frac{1}{6-2\sqrt{5}}\)
c/ ta có A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\left(x>0;x\ne1\right)\)
=\(1+\frac{2}{\sqrt{x}-1}\)
vì 1\(\in Z\) nên để A\(\in Z\) thì \(\frac{2}{\sqrt{x}-1}\in Z\)
\(\Rightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)mà x>0;x khác 1 nên \(\sqrt{x}-1>-1\)
\(\Rightarrow\sqrt{x}-1\in\left\{1;2\right\}\)
trường hợp 1: \(\sqrt{x}-1=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(TM\right)\)
trường hợp 2:\(\sqrt{x}-1=2\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\left(TM\right)\)
vậy x=4 hoặc x=9 thì A có giá trị nguyên
d/ ta có \(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}\left(x>0;x\ne1\right)\)
để A=-3 thì \(\frac{\sqrt{x}+1}{\sqrt{x}-1}=-3\Leftrightarrow\sqrt{x}+1=-3\sqrt{x}+3\)
\(\Leftrightarrow4\sqrt{x}=2\Leftrightarrow x=\frac{1}{4}\left(TM\right)\)
vậy x=1/4 thì A=-3
