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Điều kiện: $-\frac{1}{3}\le x\le6$.
Ta có:
$\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0$
$\Leftrightarrow (\sqrt{3x+1}-4)+(1-\sqrt{6-x})+3x^2-14x-5=0$
$\Leftrightarrow \frac{3x-15}{\sqrt{3x+1}+4}+\frac{x-5}{1+\sqrt{6-x}}+(x-5)(3x+1)=0$
$\Leftrightarrow (x-5)\left(\frac{3}{\sqrt{3x+1}+4}+\frac{1}{1+\sqrt{6-x}}+3x+1\right)=0$
$\Leftrightarrow x=5$, vì $\frac{3}{\sqrt{3x+1}+4}+\frac{1}{1+\sqrt{6-x}}+3x+1>0$
Vậy: $x=5$
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