Ta có: \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\)
\(=\left|x-1\right|+\left|x+2\right|\)
\(=\left|1-x\right|+\left|x+2\right|\ge\left|1-x+x+2\right|=\left|3\right|=3\)
Dấu "=" xảy ra khi: \(\left(1-x\right)\left(x+2\right)\ge0\)
\(\Rightarrow-2\le x\le1\)
Vậy \(-2\le x\le1\)
\(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=3\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}=3\)
\(\Leftrightarrow\left|x-1\right|+\left|x+2\right|=3\)(1)
Xét \(\left|x-1\right|+\left|x+2\right|\)
\(=\left|-\left(x-1\right)\right|+\left|x+2\right|\)
\(=\left|1-x\right|+\left|x+2\right|\)
\(\ge\left|1-x+x+2\right|=\left|3\right|=3\)( BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\))
Dấu "=" xảy ra ( tức (1) ) khi ab ≥ 0
=> \(\left(1-x\right)\left(x+2\right)\ge0\)
=> \(-2\le x\le1\)
Vậy \(-2\le x\le1\)là nghiệm của pt
