Nhận thấy : \(x^2+x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\Rightarrow\left|x^2+x+1\right|=x^2+x+1\)
\(x^2+3x+7=\left(x-\frac{3}{2}\right)^2+\frac{19}{4}>0\Rightarrow\left|x^2+3x+7\right|=x^2+3x+7\)
Do đó : \(A=2x^2+4x+8=2\left(x^2+2x+1\right)+6=2\left(x+1\right)^2+6\ge6\)
Dấu đẳng thức xảy ra <=> x = -1
Vậy \(MinA=6\Leftrightarrow x=-1\)
mình ko biết viết dấu giá trị tuyệt đối nên m` thấy = ngoặc vuông nhé ...
A=[\(^{x^2+2x\times\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1}\)]+ [ \(x^2+2x\cdot\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+7\)]
=[ (\(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)] + [ \(\left(x+\frac{3}{2}\right)^2+\frac{19}{4}\)]
vì \(\left(x+\frac{1}{2}\right)^2va\left(x+\frac{3}{2}\right)^2>0moix\)
\(\Rightarrow\)A=\(\left(x+\frac{1}{2}\right)^2+\left(x+\frac{3}{2}\right)^2+\frac{3}{4}+\frac{19}{4}\)
=\(\left(x+\frac{1}{2}\right)^2+\left(x+\frac{3}{2}\right)^2+\frac{22}{4}\)
vi \(\left(x+\frac{1}{2}\right)^2+\left(x+\frac{3}{2}\right)^2\ge0moix\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\left(x+\frac{3}{2}\right)^2+\frac{22}{4}\ge\frac{22}{4}\)
\(\Rightarrow A\ge\frac{22}{4}\)
=>min A=22/4
dau = xay ra \(\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)
Ta có:
\(A=\left|x^2+x+1\right|+\left|x^2+3x+7\right|\)
\(\left|x^2+x+1\right|\ge1\Leftrightarrow x=0\)
\(\left|x^2+3x+7\right|\ge7\Leftrightarrow x=0\)
\(\Rightarrow A=\left|x^2+x+1\right|+\left|x^2+3x+7\right|\ge1+7\)
\(\Rightarrow A\ge8\)
Vậy giá trị nhỏ nhất của A=8
