ĐKXĐ: \(x\ge\frac{1}{2}\)
\(pt\Leftrightarrow\sqrt{5x-1}-3+x^2+x-6=\sqrt{x+2}-2\)
\(\Leftrightarrow\frac{5\left(x-2\right)}{\sqrt{5x-1}+3}+\left(x-2\right)\left(x+3\right)=\frac{x-2}{\sqrt{x+2}+2}\)
\(\Leftrightarrow\left(\frac{5}{\sqrt{5x-1}+3}+x+3-\frac{1}{\sqrt{x+2}+2}\right)\left(x-2\right)=0\left(1\right)\)
Do \(x\ge\frac{1}{5}\Rightarrow\frac{1}{\sqrt{x+2}+2}\le\frac{5}{10+\sqrt{55}}\)
\(\Rightarrow-\frac{1}{\sqrt{x+2}+2}\ge-\frac{5}{10+\sqrt{55}}\Rightarrow x+3-\frac{1}{\sqrt{x+2}+2}>0\)
\(\Rightarrow\frac{5}{\sqrt{5x-2}+3}+x+3-\frac{1}{\sqrt{x+2}+2}>0\)
Khi đó \(\left(1\right)\Leftrightarrow x=2\left(tm\right)\)
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