Question

Giải phương trình: \(\sqrt{x^2+2x+5}+\sqrt{2x^2+4x+6}=4\)

Answer 1

\(\sqrt{x^2+2x+5}+\sqrt{2x^2+4x+6}=4\\ \Leftrightarrow\sqrt{x^2+2x+1+4}+\sqrt{2x^2+4x+2+4}=4\\ \Leftrightarrow\sqrt{\left(x^2+2x+1\right)+4}+\sqrt{2\left(x^2+2x+1\right)+4}=4\\ \Leftrightarrow\sqrt{\left(x+1\right)^2+4}+\sqrt{2\left(x+1\right)^2+4}=4\\ Do\text{ }\sqrt{\left(x+1\right)^2+4}+\sqrt{2\left(x+1\right)^2+4}=4\ge\sqrt{4}+\sqrt{4}=4\\ \text{Dấu "=" xảy ra khi }:x+1=0\\ \Leftrightarrow x=-1\)