Tìm GTNN của |2x - 2| + |x + 1|
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1.
$x(x+2)(x+4)(x+6)+8$
$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$
$=a(a+8)+8$ (đặt $x^2+6x=a$)
$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$
Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$
2.
$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$
$=5-(x^2+5x-6)(x^2+5x+6)$
$=5-[(x^2+5x)^2-6^2]$
$=41-(x^2+5x)^2\leq 41$
Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(A=\dfrac{2013}{x^2}-\dfrac{2}{x}+1=2013\left(\dfrac{1}{x}-\dfrac{1}{2013}\right)^2+\dfrac{2012}{2013}\ge\dfrac{2012}{2013}\)
Dấu "=" xảy ra khi \(x=2013\)
b.
\(B=\dfrac{4x^2+2-4x^2+4x-1}{4x^2+2}=1-\dfrac{\left(2x-1\right)^2}{4x^2+2}\le1\)
\(B_{max}=1\) khi \(x=\dfrac{1}{2}\)
\(B=\dfrac{-2x^2-1+2x^2+4x+2}{4x^2+2}=-\dfrac{1}{2}+\dfrac{\left(x+1\right)^2}{2x^2+1}\ge-\dfrac{1}{2}\)
\(B_{max}=-\dfrac{1}{2}\) khi \(x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(P=\dfrac{2x^2+x}{\left(x+1\right)^2}\Rightarrow P+\dfrac{1}{4}=\dfrac{9x^2+6x+1}{4\left(x+1\right)^2}=\dfrac{\left(3x+1\right)^2}{4\left(x+1\right)^2}\ge0\).
Dấu "=" xảy ra khi và chỉ khi \(x=-\dfrac{1}{3}\).
Vậy..
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{x^2+2x+1}{x^2-4x+5}=\dfrac{\left(x+1\right)^2}{\left(x-2\right)^2+1}\)
Do \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\\left(x-2\right)^2+1>0\end{matrix}\right.\) ;\(\forall x\)
\(\Rightarrow A\ge0\) ;\(\forall x\)
\(A_{min}=0\) khi \(x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+2x+3\)
\(=\left(x^2+2x+1\right)+2\)
\(=\left(x+1\right)^2+2\)
Do \(\left(x+1\right)^2\ge0\) với mọi x
\(\Rightarrow x^2+2x+3\ge2\)
Dấu = khi x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(x-1=t\Rightarrow x=t+1\)
\(A=\dfrac{2\left(t+1\right)^2-6\left(t+1\right)+5}{t^2}=\dfrac{2t^2-2t+1}{t^2}=\dfrac{1}{t^2}-\dfrac{2}{t}+2=\left(\dfrac{1}{t}-1\right)^2+1\ge1\)
\(A_{min}=1\) khi \(t=1\Rightarrow x=2\)
Ta có: |2x - 2| + |x + 1|
= 2.|x - 1| + |x + 1|
= |x - 1| + |x - 1| + |x + 1|
Có: \(\begin{cases}\left|x-1\right|\ge0\forall x\\\left|x-1\right|\ge1-x\\\left|x+1\right|\ge x+1\end{cases}\)
\(\Rightarrow A\ge0+\left(1-x\right)+\left(x+1\right)=2\)
Dấu "=" xảy ra khi \(\begin{cases}\left|x-1\right|=0\\\left|x-1\right|=1-x\\\left|x+1\right|=x+1\end{cases}\)\(\Rightarrow\begin{cases}x-1=0\\x-1\le0\\x+1\ge0\end{cases}\)\(\Rightarrow\begin{cases}x=1\\x\le1\\x\ge-1\end{cases}\)
=> x = 1
Vậy GTNN của |2x - 2| + |x + 1| là 2 khi x = 1