Giải bpt:

Question

Giải bpt:

$\sqrt[3]{7x-8}+1\geqslant (\sqrt{2x-1}-1)^{2}$

Lionel Messi · Answer 1 · 3/28/2016 3:11:32 PM

ĐK:

$x\geq \frac{1}{2}$

BPT

$\Leftrightarrow 2x-1-2\sqrt{2x-1}-\sqrt[3]{7x-8}\leq 0$

$\Leftrightarrow [(x+1)-2\sqrt{2x-1}]+[(x-2)-\sqrt[3]{7x-8}]\leq 0$

$\Leftrightarrow \frac{(x+1)^{2}-4(2x-1)}{x+1+2\sqrt{2x-1}}+\frac{(x-2)^{3}-(7x-8)}{(x-2)^{2}+(x-2)\sqrt[3]{7x-8}+\sqrt[3]{(7x-8)^{2}}}\leq 0$

$\Leftrightarrow \frac{(x-1)(x-5)}{...........}+\frac{x(x-1)(x-5)}{............}\leq 0$

$\Leftrightarrow (x-1)(x-5)(\frac{1}{x+1+2\sqrt{2x-1}}+\frac{x}{(x-2)^{2}+(x-2)\sqrt[3]{7x-8}+\sqrt[3]{(7x-8)^{2}}})\leq 0$

Mà

$(......)>0\Rightarrow (x-1)(x-5)\leq 0\Rightarrow 1\leq x\leq 5$

Vậy

$1\leq x\leq 5$

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