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A=\(\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
=\(\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x-2}}\)
Vậy A=\(\frac{\sqrt{x}}{\sqrt{x}-2}\)vs x\(\ge0;x\ne4\)
C=\(\left(\frac{1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\times\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{1+x}{\sqrt{x}}\)
Vậy C=\(\frac{1+x}{\sqrt{x}}\)vs x>0
1. \(N=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\frac{\sqrt{x}-3}{2\sqrt{x}-x}\)
\(N=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}+\frac{4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\frac{\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\)
\(N=\left(\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\frac{\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\)
\(N=\left(\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\frac{\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\)
\(N=\left(\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right).\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(N=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(N=\frac{4x}{x-3}\)
Vậy \(N=\frac{4x}{x-3}\)với \(x>0,x\ne4,x\ne9\)
2.Với \(x>0,x\ne4,x\ne9\)
Ta có \(N< 0\)\(\Leftrightarrow\frac{4x}{x-3}< 0\)\(\Leftrightarrow x-3< 0\)(Vì \(x>0\Leftrightarrow4x>0\)\(với\forall x\))\(\Leftrightarrow x< 3\)
Vậy ..........
3. Với \(x>0,x\ne4,x\ne9\)
Ta có \(\left|N\right|=1\Leftrightarrow\left|\frac{4x}{x-3}\right|=1\Leftrightarrow\orbr{\begin{cases}\frac{4x}{x-3}=1\\\frac{4x}{x-3}=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}4x=3-x\\4x=x-3\end{cases}}\)\(\orbr{\begin{cases}x=\frac{3}{5} \left(N\right)\\x=-1\left(N\right)\end{cases}}\)
Vậy ...............
\(A=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+8\sqrt{x}}{x-4}:\frac{2\left(\sqrt{x}+2\right)-2\sqrt{x}-3}{\sqrt{x}+2}\)
\(A=\frac{2x}{x-4}.\left(\sqrt{x}+2\right)=\frac{2x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{2x}{\sqrt{x}-2}\)