Ta có: \(B=\left(x-3\right)^2+\left(x+1\right)^2\)
\(=x^2-6x+9+x^2+2x+1\)
\(=2x^2-4x+10\)
\(=2\left(x^2-2x+1+4\right)\)
\(=2\left(x-1\right)^2+8\ge8\forall x\)
Dấu '=' xảy ra khi x-1=0
hay x=1
Ta có B=\(\left(x-3\right)^2+\left(x+1\right)^2=x^2-6x+9+x^2+2x+1\)
\(=2x^2-4x+10=2\left(x^2-2x+5\right)\)
\(=2\left(x^2-2x+1+4\right)\)
=\(2\left(x^2-2x+1\right)+8\) =\(2\left(x-1\right)^2+8\)
Ta có \(2\left(x-1\right)^2\ge0\) \(\Rightarrow2\left(x-1\right)^2+8\ge8\)
Vậy Min A=8 \(\Leftrightarrow x=1\)
B=(x−3)2+(x+1)2B=(x−3)2+(x+1)2
=x2−6x+9+x2+2x+1=x2−6x+9+x2+2x+1
=2x2−4x+10=2x2−4x+10
=2(x2−2x+1+4)=2(x2−2x+1+4)
=2(x−1)2+8≥8∀x=2(x−1)2+8≥8∀x
Dấu '=' xảy ra khi x-1=0
hay x=1
B=(x−3)2+(x+1)2B=(x−3)2+(x+1)2
=x2−6x+9+x2+2x+1=x2−6x+9+x2+2x+1
=2x2−4x+10=2x2−4x+10
=2(x2−2x+1+4)=2(x2−2x+1+4)
=2(x−1)2+8≥8∀x=2(x−1)2+8≥8∀x
