anakim111

7) ĐK:$\frac{5}{6}\leq x\leq 6$

$x^{2}-2x-5-\sqrt{6-x}-\sqrt{2x+6}-\sqrt{6x-5}\leq 0$

$2(x-5)(10x+13)+20(1-\sqrt{6-x})+5\sqrt{2x+6}(\sqrt{2x+6}-4)+4\sqrt{6x-5}(\sqrt{6x-5}-5)\leq 0$

$\Leftrightarrow (x-5)[2(10x+13)+\frac{20}{1+\sqrt{6-x}}+\frac{10\sqrt{2x+6}}{\sqrt{2x+6}+4}+\frac{24\sqrt{6x-5}}{\sqrt{6x-5}+5}]\leq 0$

$\Leftrightarrow x\leq 5$

Vậy $\frac{5}{6}\leq x\leq 5$

Truy ngược dấu! hay quá bác!