Lời giải:
a) ĐK: $x\geq -2$
PT \(\Leftrightarrow \sqrt{(x+2)-4\sqrt{x+2}+4}+\sqrt{(x+2)-6\sqrt{x+2}+9}=1\)
\(\Leftrightarrow \sqrt{(\sqrt{x+2}-2)^2}+\sqrt{(\sqrt{x+2}-3)^2}=1\)
\(\Leftrightarrow |\sqrt{x+2}-2|+|\sqrt{x+2}-3|=1\)
Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:
\(|\sqrt{x+2}-2|+|\sqrt{x+2}-3|=|\sqrt{x+2}-2|+|3-\sqrt{x+2}|\)
\(\geq |\sqrt{x+2}-2+3-\sqrt{x+2}|=1\)
Dấu "=" xảy ra khi $(\sqrt{x+2}-2)(3-\sqrt{x+2})\geq 0$
$\Leftrightarrow 3\geq \sqrt{x+2}\geq 2$
$\Leftrightarrow 7\geq x\geq 2$
Vậy.........
b)
ĐK: $x\geq \frac{5}{2}$
PT $\Leftrightarrow \sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=14$
$\Leftrightarrow \sqrt{(2x-5)+2\sqrt{2x-5}+1}+\sqrt{(2x-5)+6\sqrt{2x-5}+9}=14$
$\Leftrightarrow \sqrt{(\sqrt{2x-5}+1)^2}+\sqrt{(\sqrt{2x-5}+3)^2}=14$
$\Leftrightarrow \sqrt{2x-5}+1+\sqrt{2x-5}+3=14$
$\Leftrightarrow \sqrt{2x-5}=5$
$\Rightarrow x=15$ (tm)
