Question

[đanh khoa]

\(\sqrt{x+2-3\sqrt{2x-5}}+\sqrt{x-2+\sqrt{2x-5}}=2\sqrt{2}\)GIẢI PHƯƠNG TRÌNH

Answer 1

[Hoàng Hà]

ĐKXĐ: \(2x-5\ge0\Leftrightarrow x\ge2,5\)

pt\(\Leftrightarrow\sqrt{2x+4-2.3\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)\(\Leftrightarrow\sqrt{2x-5-2.3\sqrt{2x-5}+9}+\sqrt{2x-5+2\sqrt{2x-5}+1}=4\)

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

\(\Leftrightarrow\left|\sqrt{2x-5}-3\right|+\left|\sqrt{2x-5}+1\right|=4\)

\(\Leftrightarrow\left|3-\sqrt{2x-5}\right|+\left|\sqrt{2x-5}+1\right|=4\)

Có: \(VT=\left|3-\sqrt{2x-5}\right|+\left|\sqrt{2x-5}+1\right|\ge\left|3-\sqrt{2x-5}+\sqrt{2x-5}+1\right|=4=VP\)

Dấu "=" xảy ra khi \(\left(3-\sqrt{2x-5}\right)\left(\sqrt{2x-5}+1\right)\ge0\)

Mà \(\sqrt{2x-5}+1\ge0\Rightarrow3-\sqrt{2x-5}\ge0\Rightarrow\sqrt{2x-5}\le3\)

\(\Rightarrow0\le\sqrt{2x-5}\le3\)

\(\Leftrightarrow0\le2x-5\le9\)

\(\Leftrightarrow2,5\le x\le7\)(TM)