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$3\sqrt{2(3x+4)^{3}}\leq (x^{2}+2x-2)\sqrt{x^{2}+2x-3}+(19x+26).\sqrt{x+1}$

๖ۣۜColdღ · Answer 1 · 5/22/2016 3:56:05 PM

ĐK;

$x\geq-1$

pt

$\Leftrightarrow (9x+12)\sqrt{6x+8}-(19x+26)\sqrt{x+1} \leq \sqrt{(x^{2}+2x-3)^{3}} +\sqrt{x^{2}+2x-3}$ (1)

Đặt

$\sqrt{6x+8}=a;\sqrt{x+1}=b \Rightarrow 9x+12=a^{2}+3b^{2}+1;19x+26=3a^{2}+b^{2}+1$

(1)

$\Leftrightarrow (a^{2}+3b^{2}+1)a-(3a^{2}+b^{2}+1)b\leq \sqrt{(x^{2}+2x-3)^{3}}+\sqrt{x^{2}+2x-3}$

$\Leftrightarrow (a-b)^{3}+(a-b)\leq \sqrt{(x^{2}+2x-3)^{3}}+\sqrt{x^{2}+2x-3}$

$\Leftrightarrow (a-b-\sqrt{x^{2}+2x-3})(a-b+(a-b)\sqrt{x^{2}+2x-3}+x^{2}+2x-3+1)$

$\Leftrightarrow a-b\leq \sqrt{x^{2}+2x-3}$ do (....)>0

$\Leftrightarrow \sqrt{6x+8} -\sqrt{x+1}\leq \sqrt{x^{2}+2x-3}$

$\Leftrightarrow 6x+8\leq (\sqrt{x^{2}+2x-3} +\sqrt{x+1})^{2}$

$\Leftrightarrow (\sqrt{x^{2}-1}-\sqrt{x+3})(3\sqrt{x+3}+\sqrt{x^{2}-1})\geq0$

$\Leftrightarrow x^{2}-1 \geq x+3$

$\Leftrightarrow x \leq \frac{1-\sqrt{17}}{2}or x\geq\frac{1+\sqrt{17}}{2}$

kh vs ĐK

$\Rightarrow x\geq \frac{1+\sqrt{17}}{2}$

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