\(A=x^2-4x+7\)
\(A=x^2-4x+4+3\)
\(A=\left(x-2\right)^2+3\)
Vậy \(MIN_A=3\)
Dấu = xảy ra khi \(\left(x-2\right)^2=0\Rightarrow x=2\)
\(\text{a) }A=x^2-4x+7\\ A=x^2-2.x.2+2^2+3\\ A=\left(x^2-2.x.2+2^2\right)+3\\ A=\left(x-2\right)^2+3\\ \text{Ta có : }\left(x-2\right)^2\ge0\\ \Rightarrow A=\left(x-2\right)^2+3\ge3\\ \text{Dấu }"="\text{xảy ra khi: }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{Vậy }A_{\left(Min\right)}=3\text{ xảy ra khi: }x=2\\ \)
\(\text{b) }B=x^2+8x\\ B=x^2+2\cdot x\cdot4+16-16\\ B=\left(x^2+2\cdot x\cdot4+4^2\right)-16\\ B=\left(x+4\right)^2-16\\ \text{Ta có : }\left(x+4\right)^2\ge0\\ \Rightarrow B=\left(x+4\right)^2-16\ge-16\\ \text{ Dấu }"="\text{ xảy ra khi: }\\ \left(x+4\right)^2=0\\ \Leftrightarrow x+4=0\\ \Leftrightarrow x=-4\\ \text{ Vậy }B_{\left(Min\right)}=-16\text{ khi }x=-4\\ \)
\(\text{c) }C=2x^2+4x+15\\ C=\left(2x^2+4x+2\right)+13\\ C=2\left(x^2+2x+1\right)+13\\ C=2\left(x^2+2x+1^2\right)+13\\ C=2\left(x+1\right)^2+13\\ \text{Ta có : }\left(x+1\right)^2\ge0\\ \Rightarrow2\left(x+1\right)^2\ge0\\ \Rightarrow C=2\left(x+1\right)^2+13\ge13\\ \text{ Dấu }"="\text{ xảy ra khi: }\\ 2\left(x+1\right)^2=0\\ \Leftrightarrow\left(x+1\right)^2=0\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\\ \text{Vậy }C_{\left(Min\right)}=13\text{ khi }x=-1\)
