Question

Giải phương trình \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)

Answer 1

Lời giải:
ĐK: $x\geq \frac{5}{2}$

PT $\Leftrightarrow \sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4$

$\Leftrightarrow \sqrt{(2x-5)+6\sqrt{2x-5}+9}+\sqrt{(2x-5)-2\sqrt{2x-5}+1}=4$

$\Leftrightarrow \sqrt{(\sqrt{2x-5}+3)^2}+\sqrt{(\sqrt{2x-5}-1)^2}=4$

$\Leftrightarrow |\sqrt{2x-5}+3|+|1-\sqrt{2x-5}|=4$

Theo BĐT dạng $|a|+|b|\geq |a+b|$ thì:

$|\sqrt{2x-5}+3|+|1-\sqrt{2x-5}|\geq |\sqrt{2x-5}+3+1-\sqrt{2x-5}|=4$

Dấu "=" xảy ra khi $(\sqrt{2x-5}+3)(1-\sqrt{2x-5})\geq 0$

$\Leftrightarrow 1-\sqrt{2x-5}\geq 0$

$\Leftrightarrow 0\leq 2x-5\leq 1$

$\Leftrightarrow \frac{5}{2}\leq x\leq 3$