Đáp án:
a. \(A_{\min}=\dfrac 74\Leftrightarrow x=-\dfrac 52\)
b. \(B_{\min}=1\Leftrightarrow x=-2\)
Giải thích các bước giải:
a. Ta có
\(\begin{split}A&=x^2+5x+8\\ &=x^2+2\cdot \dfrac 52x+\dfrac{25}{4}+\dfrac 74\\ &=\left(x+\dfrac52\right)^2+\dfrac 74\ge \dfrac 74\ \left(\text{do} \ \left(x+\dfrac52\right)^2\ge0\Rightarrow\left(x+\dfrac52\right)^2+\dfrac 74\ge \dfrac 74 \right)\end{split}\)
Suy ra: \(A_{\min}=\dfrac 74\Leftrightarrow \left(x+\dfrac52\right)^2=0\Leftrightarrow x=-\dfrac 52\)
b. Ta có
\(\begin{split}B&=2x^2+8x+9\\ &=2\left(x^2+4x+4\right)+1\\ &=2\left(x+2\right)^2+1\ge 1\ \left(\text{do}\ 2(x+2)^2\ge0\Rightarrow2\left(x+2\right)^2+1\ge 1 \right)\end{split}\)
Suy ra: \(B_{\min}=1\Leftrightarrow 2\left(x+2\right)^2=0\Leftrightarrow x=-2\)
