Tìm GTNN của \(A=|x|+|2x+1|+|3x+2|+...+|2011x+2010|\)
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a) x+2x+3x+4x+...+2011x = 2012.2013
\(\Rightarrow\) x(1+2+3+4+...+2011) = 4050156
\(\Rightarrow\) x.2023066 = 4050156
\(\Rightarrow\) x = 4026/2011
\(x+2x+3x+...+2011x=2012.1013\)
\(\dfrac{2011\left(2011+1\right)}{2}x=2012.2013\)
\(x=2012.2013.\dfrac{2}{2011.2012}\)
\(x=\dfrac{4026}{2011}\)
\(b,ĐKXĐ:x>0\)
\(D=2011\sqrt{x}-2+\frac{1}{\sqrt{x}}\)\(=2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\)
Áp dụng bđt Cauchy cho 2 số dương \(2011\sqrt{x}\)và\(\frac{1}{\sqrt{x}}\)ta được:
\(2011\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{2011\sqrt{x}.\frac{1}{\sqrt{x}}}\)
\(\Leftrightarrow2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\ge2\sqrt{2011}-2\)
\(\Leftrightarrow D\ge2\sqrt{2011}-2\)
Dấu "=" xảy ra \(\Leftrightarrow2011\sqrt{x}=\frac{1}{\sqrt{x}}\Leftrightarrow x=\frac{1}{2011}\left(TMĐK\right)\)
Áp dụng BĐT AM-GM ta có:
\(A=\frac{2011x+2012\sqrt{1-x^2}+2013}{\sqrt{1-x^2}}\)\(=\frac{2011x+2013}{\sqrt{1-x^2}}+2012\)
\(=\frac{2012\left(x+1\right)+\left(1-x\right)}{\sqrt{1-x^2}}+2012\)\(\ge\frac{2\sqrt{2012\left(x+1\right)\left(1-x\right)}}{\sqrt{1-x^2}}+2012\)
\(\ge\frac{2\sqrt{2012\left(1-x^2\right)}}{\sqrt{1-x^2}}+2012=2\sqrt{2012}+2012\)
Ta có :
x = 2012
x - 1 = 2011
P(x) = x2012 - 2011x2011 - 2011x2010 - .... - 2011x2 - 2011x - 1
P(x) = x2012 - (x - 1)x2011 - (x - 1)x2010 - ..... - (x - 1)x2 - (x - 1)x - 1
P(x) = x2012 - x2012 + x2011 - x2011 + x2010 - ...... - x3 + x2 - x2 + x - 1
P(x) = x - 1
P(2012) = 2012 - 1 = 2011
\(x+2x+3x+....+2001x=2012.2013\)
\(\Leftrightarrow x\left(1+2+3+...+2001\right)=2012.2013\)
\(\Leftrightarrow x.\frac{\left(2011+1\right).2011}{2}=2012.2013\)
\(\Leftrightarrow x=\frac{2012.2013.2}{\left(2011+1\right).2011}\)
\(\Leftrightarrow x=\frac{2.2013}{2011}\)
X+2X+3X+4X+...+2011X=2012.2013
<=> X(1+2+3+4+......+2011)=2012.2013
<=>2023066X=4050156
<=>X=4050156:2023066=\(\frac{4026}{2011}\)
Vậy x=\(\frac{4026}{2011}\)