Question

Tìm min 

A=(x-2)^2+|x-1|+5

B=2(x+1)^2-|x+3|-11

mình cần gấp

Answer 1

Lời giải:

$A=(x-2)^2+|x-1|+5$

Nếu $x\geq 1$ thì:

$A=(x-2)^2+x-1+5=x^2-4x+4+x-1+5=x^2-3x+8=(x-\frac{3}{2})^2+\frac{23}{4}\geq \frac{23}{4}(*)$
Nếu $x< 1$:

$A=(x-2)^2+1-x+5=x^2-5x+10=(x-1)(x-4)+6> 6(**)$

Từ $(*); (**)\Rightarrow A_{\min}=\frac{23}{4}$ khi $x=\frac{3}{2}$
 

 

Answer 2

Lời giải:
\(B=2(x+1)^2-|x+3|-11\)

Nếu $x\geq -3$ thì:

\(B=2(x+1)^2-(x+3)-11=2x^2+3x-12=2(x+\frac{3}{4})^2-\frac{105}{8}\)

\(\geq \frac{-105}{8}\) (1)

Nếu $x< -3$
$B=2(x+1)^2+(x+3)-11=2x^2+5x-6=(x+3)(2x+1)-9> -9$ (2)

Từ $(1); (2)\Rightarrow B_{\min}=\frac{-105}{8}$ khi $x+\frac{3}{4}=0\Leftrightarrow x=\frac{-3}{4}$