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Vì \(\left\{{}\begin{matrix}\left|2x-27\right|^{2011}\text{≥0,∀x}\\\left(3y+10\right)^{2012}\text{≥0,∀y}\end{matrix}\right.\)
⇒ \(\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\text{≥0,∀x},y\)
Dấu "=" ⇔ \(\left\{{}\begin{matrix}2x-27=0\\3y+10=0\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=\dfrac{27}{2}\\y=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy ...
\(f\left(0\right)=2010\Rightarrow a.0^2+b.0+c=2010\Rightarrow c=2010\)
\(f\left(1\right)=2011\Rightarrow a.1^2+b.1+c=2011\Rightarrow a+b+c=2011\)
\(\Rightarrow a+b+2010=2011\Rightarrow a+b=1\) (1)
\(f\left(-1\right)=2012\Rightarrow a.\left(-1\right)^2+b.\left(-1\right)+c=2012\)
\(\Rightarrow a-b+c=2012\Rightarrow a-b+2010=2012\)
\(\Rightarrow a-b=2\Rightarrow a=b+2\)
Thế vào (1) \(\Rightarrow b+2+b=1\Rightarrow2b=-1\Rightarrow b=-\dfrac{1}{2}\)
\(\Rightarrow a=b+2=-\dfrac{1}{2}+2=\dfrac{3}{2}\)
\(\Rightarrow f\left(x\right)=\dfrac{3}{2}x^2-\dfrac{1}{2}x+2010\)
\(\Rightarrow f\left(-2\right)=\dfrac{3}{2}.\left(-2\right)^2-\dfrac{1}{2}.\left(-2\right)+2010=2017\)
Ta có : \(\frac{x+1}{2013}+\frac{x+2}{2012}+\frac{x+3}{2011}=-3.\)
\(\Leftrightarrow\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=-3+3\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)
Mà \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\ne0\) nên \(x+2014=0\Leftrightarrow x=-2014\)
Vây \(x=-2014\)
\(D=-x^2+8x-4\)
\(D=-x^2+8x-16+12\)
\(D=-\left(x-4\right)^2+12\)
Có \(-\left(x-4\right)^2\le0\)
\(\Rightarrow D\le12\)
Vậy Max D = 12<=>x=4
\(E=-2x^2-4x+5\)
\(E=-2x^2-4x-2+7\)
\(E=-2\left(x+1\right)^2+7\le7\)
Vậy Max E = 7<=>x=-1
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Link:
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Tìm GTNN nha mọi ng,e ghi nhầm
\(A=\left|x-2011\right|+\left|x-2012\right|\ge\left|x-2011+2012-x\right|=1\)
Dấu "=" <=> 2011 < x < 2012
Vậy Amax = 1 <=> 2011 < x < 2012