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ĐK: $\frac{-1}{3}\leq x\leq 6$
$PT\Leftrightarrow (\sqrt{3x+1}-4)+(1-\sqrt{6-x})+(3x^2-14x-5)=0$
$\Leftrightarrow \frac{3(x-5)}{\sqrt{3x+1}+4}+\frac{x-5}{1+\sqrt{6-x}}+(3x+1)(x-5)=0$
$\Leftrightarrow (x-5)[\frac{3}{\sqrt{3x+1}+4}+\frac{1}{\sqrt{6-x}+1}+3x+1]=0$
Mà $[....]>0\Rightarrow x-5=0\Leftrightarrow x=5$
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