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\(Q=\left|x-2010\right|+\left(y+2011\right)^{2010}+2011\)
Ta có:\(\hept{\begin{cases}\left|x-2010\right|\ge0\\\left(y+2011\right)^{2010}\ge0\end{cases}}\)
Nên \(\left|x-2010\right|+\left(y+2011\right)^{2010}+2011\ge2011\)
Vậy \(Q_{min}=2011\Leftrightarrow\hept{\begin{cases}x-2010=0\\y+2011=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2010\\y=-2011\end{cases}}\)
A=|x-2008|+|2009-x|+|y-2010|+|x-2011|+2011
≥|x-2008+2009-x|+|y-2010|+|x-2011|+2011
= |y-2010|+|x-2011|+2012≥2012
Dấu = xảy ra khi : {y−2010=0x−2011=0{y−2010=0x−2011=0
<=> {y=2010x=2011{y=2010x=2011
Vay GTNN cua A=2012 khi {x=2011;y=2010
Vì |x-2010|\(\ge\)0
(y+2011) 2010\(\ge\)0
=>|x-2010|+(y+2011) 2010\(\ge\)0
=>A=|x-2010| + (y+2011) 2010 +2011 \(\ge\)0+2011
Dấu "=" xảy ra khi |x-2010|=(y+2011)2010=0
<=>x=2010 và y=-2011
Vậy Amin=2011 khi x=2010 và y=-2011
Lời giải:
Ta thấy:
\(|x-2010|\geq 0, \forall x\in\mathbb{R}\)
\((y+2011)^{2010}=[(y+2010)^{1005}]^2\geq 0, \forall y\in\mathbb{R}\)
\(\Rightarrow A=|x-2010|+(y+2011)^{2010}+2011\geq 0+0+2011=2011\)
Vậy GTNN của $A$ là $2011$.
Dấu "=" xảy ra khi \(\left\{\begin{matrix} x-2010=0\\ y+2011=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2010\\ y=-2011\end{matrix}\right.\)
a: \(\left(x-2\right)^2>=0\)
\(\left|y-x\right|>=0\)
Do đó: \(\left(x-2\right)^2+\left|y-x\right|>=0\forall x,y\)
=>\(\left(x-2\right)^2+\left|y-x\right|+3>=3\forall x,y\)
=>A>=3 với mọi x,y
Dấu = xảy ra khi x-2=0 và y-x=0
=>x=2=y
b: \(\left|x+5\right|>=0\)
=>\(\left|x+5\right|+5>=5\)
=>B>=5 với mọi x
Dấu = xảy ra khi x+5=0
=>x=-5
c: \(\left|x-2010\right|>=0\)
=>\(-\left|x-2010\right|< =0\)
=>\(-\left|x-2010\right|+2012< =2012\)
=>\(C=\dfrac{2011}{2012-\left|x-2010\right|}>=\dfrac{2011}{2012}\forall x\)
Dấu = xảy ra khi x=2010
a) Ta có:
\(A=\left(x-2\right)^2+\left|y-x\right|+3\)
Mà: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left|y-x\right|\ge0\end{matrix}\right.\)
\(\Rightarrow A=\left(x-2\right)^2+\left|y-x\right|+3\ge3\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}x-2=0\\y-x=0\end{matrix}\right.\)
\(\Rightarrow x=y=2\)
Vậy: \(A_{min}=3\Leftrightarrow x=y=2\)
b) Ta có:
\(B=\left|x+5\right|+5\)
Mà: \(\left|x+5\right|\ge0\)
\(\Rightarrow B=\left|x+5\right|+5\ge5\)
Dấu "=" xảy ra:
\(x+5=0\Rightarrow x=-5\)
Vậy: \(B_{min}=5\Leftrightarrow x=-5\)
c) Ta có:
\(C=\dfrac{2011}{2012-\left|x-2010\right|}\)
Mà: \(\left|x-2010\right|\ge0\)
\(\Rightarrow C=\dfrac{2011}{2012-\left|x-2010\right|}\ge\dfrac{2011}{2012}\)
Dấu "=" xảy ra khi:
\(x-2010=0\Rightarrow x=2010\)
Vậy: \(C_{min}=\dfrac{2011}{2012}\Leftrightarrow x=2010\)
A=/x-2008/+/2009-x/+/y-2010/+/x-2011/+2011
≥/x-2008+2009-x/+/y-2010/+/x-2011/+2011
= /y-2010/+/x-2011/+2012≥2012
Dau bang xay ra khi : \(\left\{{}\begin{matrix}y-2010=0\\x-2011=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}y=2010\\x=2011\end{matrix}\right.\)
Vay GTNN cua A=2012 khi \(\left\{{}\begin{matrix}x=2011\\y=2010\end{matrix}\right.\)
ta thấy: \(\left|x-2010\right|\ge0\); \(\left(y+2011\right)^{2020}\ge0\)
\(\Rightarrow\left|x-2010\right|+\left(y+2011\right)^{2020}+2011\ge2011\)
dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-2010=0\\y+2011=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2010\\y=-2011\end{matrix}\right.\)
vậy MinA=2011 khi\(\left\{{}\begin{matrix}x=2010\\y=-2011\end{matrix}\right.\)