có sai đề k bạn nếu sai mk xin sửa
đkxđ \(x\ne1;x\ge0\)
a,\(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{x-1}\)
\(=\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-3\sqrt{x}-1}{x-1}\)
\(=\frac{2x+3-3\sqrt{x}-1}{x-1}=\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b, khi x = \(4+2\sqrt{3}\)
\(A=\frac{2\left(3+2\sqrt{3}\right)+1}{4+2\sqrt{3}-2}=\frac{9+4\sqrt{5}}{3+2\sqrt{3}}\)
c, để A = 1/2
\(\frac{2\sqrt{x}+1}{\sqrt{x}-1}=\frac{1}{2}\Leftrightarrow4\sqrt{x}+2=\sqrt{x}-1\Leftrightarrow3\sqrt{x}=3\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(ktmđk\right)\)
d, để A<1\(\Leftrightarrow\frac{2\sqrt{x}+1}{\sqrt{x}-1}< 1\left(1\right)\)
th1 với \(\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
Ta có (1)\(\Leftrightarrow2\sqrt{x}+1< \sqrt{x}-1\Leftrightarrow\sqrt{x}< -2\left(loại\right)\)
th2 \(\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow0\le x< 1\)
ta có (1) \(\Leftrightarrow2\sqrt{x}+1>\sqrt{x}-1\Leftrightarrow\sqrt{x}>-2\left(luônđúng\right)\)
\(\Rightarrow0\le x< 1\)
e, \(A=\frac{2\sqrt{x}+1}{\sqrt{x}-1}=\frac{2\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}=2+\frac{3}{\sqrt{x}-1}\)
để A là số nguyen thì \(\frac{3}{\sqrt{x}-1}\)cũng là số nguyên
\(\left[{}\begin{matrix}\sqrt{x}-1=1\\\sqrt{x}-1=-1\\\sqrt{x}-1=3\\\sqrt{x}-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=0\\\sqrt{x}=4\\\sqrt{x}=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tmđk\right)\\x=0\left(tmđk\right)\\x=16\left(tmđk\right)\end{matrix}\right.\)