Question

\(\sqrt{5+x-4\sqrt{x+1}}+\sqrt{10+x-6\sqrt{x+1}}=1\) .giải pt trên.

Answer 1

Lời giải:
ĐKXĐ: \(x\geq -1\)

\(PT\Leftrightarrow \sqrt{(x+1)-4\sqrt{x+1}+4}+\sqrt{(x+1)-6\sqrt{x+1}+9}=1\)

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

\(\Leftrightarrow |\sqrt{x+1}-2|+|3-\sqrt{x+1}|=1\)

Áp dụng BĐT dạng $|a|+|b|\ge |a+b|$ ta có:

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

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

$\Leftrightarrow 2\leq \sqrt{x+1}\leq 3$

$\Leftrightarrow 3\leq x\leq 8$

Vậy.........

Answer 2

Lời giải:
ĐKXĐ: \(x\geq -1\)

\(PT\Leftrightarrow \sqrt{(x+1)-4\sqrt{x+1}+4}+\sqrt{(x+1)-6\sqrt{x+1}+9}=1\)

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

\(\Leftrightarrow |\sqrt{x+1}-2|+|3-\sqrt{x+1}|=1\)

Áp dụng BĐT dạng $|a|+|b|\ge |a+b|$ ta có:

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

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

$\Leftrightarrow 2\leq \sqrt{x+1}\leq 3$

$\Leftrightarrow 3\leq x\leq 8$

Vậy.........