Mình làm ở bài trước rồi nhé -..-
\(A=x^2-4x+7\)
\(=x^2-2x2+2^2+3\)
\(=\left(x-2\right)^2+3\ge3\forall x\)
Dấu"="xảy ra khi \(\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy \(Min_A=3\Leftrightarrow x=2\)
\(B=2x^2-6x\)
\(=2\left(x^2-3x\right)\)
\(=2\left[x^2-2x\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{9}{2}\right]\)
\(=2\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\right]\ge-\frac{9}{2}\forall x\)
Dấu"="xảy ra khi\(2\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)
Vậy \(Min_B=\frac{-9}{2}\Leftrightarrow x=\frac{3}{2}\)
\(C=-2x^2+8x-15\)
\(=-2\left(x^2-4x+4\right)-7\)
\(=-2\left(x-2\right)^2-7\le-7\forall x\)
Dấu"="xảy ra khi \(-2\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy \(Max_C=-7\Leftrightarrow x=2\)
