Đặt \(B=2^{2011}+2^{2010}+\cdots+2+1\)

=>2B=\(2^{2012}+2^{2011}+\cdots+2^2+2\)

=>2B-B=\(2^{2012}+2^{2011}+\cdots+2^2+2-2^{2011}-2^{2010}-\cdots-2-1\)

=>\(B=2^{2012}-1\)

Ta có: \(A=2^{2012}-\left(2^{2011}+2^{2010}+\cdots+2+1\right)\)

\(=2^{2012}-\left(2^{2012}-1\right)\)

=1

f(0)=2010

=>\(a\cdot0^2+b\cdot0+c=2010\)

=>c=2010

=>\(f\left(x\right)=a\cdot x^2+bx+2010\)

\(f\left(1\right)=2011\)

=>\(a\cdot1^2+b\cdot1+2010=2011\)

=>a+b=1

f(-1)=2012

=>\(a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+2010=2012\)

=>a-b=2

mà a+b=1

nên a=(2+1)/2=1,5; b=1-1,5=-0,5

=>f(x)=1,5x^2-0,5x+2010

\(f\left(-2\right)=1,5\cdot\left(-2\right)^2-0,5\cdot\left(-2\right)+2010\)

=6+1+2010

=2017

Ta có:

\(\frac{x+4}{2008}+1+\frac{x+3}{2009}+1=\frac{x+2}{2010}+1+\frac{x+1}{2011}+1\)

\(\frac{x+2012}{2008}+\frac{x+2012}{2009}=\frac{x+2012}{2010}+\frac{x+2012}{2011}\)

\(\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)

\(x=-2012\)

;Ko tồn tại nghiệm số thực OK

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 \(x=-2014\)

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