Question

Giải phương trình

\(\sqrt{x+3-4\sqrt{x-1}}\)

+ \(\sqrt{x+8+6\sqrt{x-1}}\)=5

Answer 1

ĐK : \(x\ge1\)

\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)

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

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

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

Ta có : \(\left|\sqrt{x-1}-2\right|=\left|2-\sqrt{x-1}\right|\)

Áp dụng BĐT \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

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

Dấu \("="\) hiển nhiên xảy ra khi \(1\le x\le5\)