A=\(\sqrt{x^2-2x+1}+\sqrt{x^2+6x+9}=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+3\right)^2}\)=|x-1|+|x+3|=|1-x|+|x+3|
Áp dụng bđt |a|+|b|\(\ge\)|a+b| ta được: A=|1-x|+|x+3|\(\ge\)|1-x+x+3|=4
Dấu "=" xảy ra khi (1-x)(x+3)\(\ge\)0 <=> \(-3\le x\le1\)
Vậy Amin=4 khi \(-3\le x\le1\)
A = \(\sqrt{x^2-2x+1}+\sqrt{x^2+6x+9}\)
= \(\sqrt{\left(1-x\right)^2}+\sqrt{\left(x+3\right)^2}\)
= 1 - x + x + 3
= 4
\(\sqrt{x^2-2x+1}\)\(+\sqrt{x^2+6x+9}\)
\(=\sqrt{\left(x+1\right)^2}+\sqrt{\left(x+3\right)^2}\)
\(=x+1+x+3\)
=2x+4
A đạt GTNN khi A=0
--> 2x+4=0
<=>2x=-4
<=> x=-2
vậy x=-2 thì A đạt GTNN
