Câu hỏi của Nguyễn Thảo Hân

Question

giải bpt :

$\frac{\sqrt{2\left(x^2+7x+3\right)}-\sqrt{x^2+x-6}-3\sqrt{x+1}}{1-2\sqrt{x^2-x+1}}\ge0$

Nguyễn Việt Lâm · Accepted Answer

ĐKXĐ: $x\ge2$

Khi đó ta có $x^2-x+1\ge3\Rightarrow1-2\sqrt{x^2-x+1}< 0$

Do đó BPT tương đương:

$\sqrt{2\left(x^2+7x+3\right)}-\sqrt{x^2+x-6}-3\sqrt{x+1}\le0$

$\Leftrightarrow\sqrt{2x^2+14x+6}\le\sqrt{x^2+x-6}+3\sqrt{x+1}$

$\Leftrightarrow2x^2+14x+6\le x^2+10x+3+6\sqrt{\left(x+1\right)\left(x^2+x-6\right)}$

$\Leftrightarrow x^2+4x+3\le6\sqrt{\left(x+1\right)\left(x+3\right)\left(x-2\right)}$

$\Leftrightarrow\left(x+1\right)\left(x+3\right)\le6\sqrt{\left(x+1\right)\left(x+3\right)\left(x-2\right)}$

$\Leftrightarrow\sqrt{\left(x+1\right)\left(x+3\right)}\le6\sqrt{x-2}$

$\Leftrightarrow\left(x+1\right)\left(x+3\right)\le36\left(x-2\right)$

$\Leftrightarrow x^2-32x+75\le0$

$\Rightarrow16-\sqrt{181}\le x\le16+\sqrt{181}$

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