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5/
Đặt \(\left\{{}\begin{matrix}\sqrt{2x-\frac{3}{x}}=a\ge0\\\sqrt{\frac{6}{x}-2x}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=\frac{3}{x}\)
Pt trở thành:
\(a-1=\frac{a^2+b^2}{2}-b\)
\(\Leftrightarrow a^2+b^2-2a-2b+2=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x-\frac{3}{x}}=1\\\sqrt{\frac{6}{x}-2x}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-x-3=0\\2x^2+x-6=0\end{matrix}\right.\) \(\Rightarrow x=\frac{3}{2}\)
4/
ĐKXĐ: \(x\ge\frac{1}{5}\)
\(\Leftrightarrow\frac{4x-3}{\sqrt{5x-1}+\sqrt{x+2}}=\frac{4x-3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-3=0\Rightarrow x=\frac{3}{4}\\\sqrt{5x-1}+\sqrt{x+2}=5\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{5x-1}-3+\sqrt{x+2}-2=0\)
\(\Leftrightarrow\frac{5\left(x-2\right)}{\sqrt{5x-1}+3}+\frac{x-2}{\sqrt{x+2}+2}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{5}{\sqrt{5x-1}+3}+\frac{1}{\sqrt{x+2}+2}\right)=0\)
\(\Leftrightarrow x=2\)
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 ...............