Question

giải pt ạ

\(\sqrt{x+2+2\sqrt{x+1}}+\sqrt{x+10-6\sqrt{x+1}}=2\sqrt{x+2-2\sqrt{x+1}}\)

Answer 1

ĐKXĐ: \(x\ge-1\)

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

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

Áp dụng BĐT trị tuyệt đối:

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

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

\(\Leftrightarrow\sqrt{x+1}-3\ge0\)

\(\Leftrightarrow x+1\ge9\)

\(\Leftrightarrow x\ge8\)