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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
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x^4-2012(x^3-x^2+x-1)
mà 2012=x
suy ra h(2012)=x^4-x.x^3+x.x^2-x.x+2012
=x^4-x^4+x^3-x^2+x
=x^3-x^2+x
=2012(2012^2-2012+1)
=2012(2012.2011+1)
=2012^2.2011+2012
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Ta có: \(x=2011\Rightarrow x+1=2012\)
Khi đó, ta có:
\(H\left(x\right)=x^4-\left(x+1\right).x^3+\left(x+1\right).x^2-\left(x+1\right).x+2012\)
\(=x^4-x^4-x^3+x^3+x^2-x^2-x+2012\)
\(\Rightarrow H\left(2011\right)=-2011+2012=1\).
Vậy \(H\left(2011\right)=1\)
Cách 2:
\(H\left(x\right)=x^4-2012x^3+2012x^2-2012x+2012\)
\(=x^4-2011x^3-x^3+2011x^2+x^2-2011x-x+2011+1\)
\(=x^3\left(x-2011\right)-x^2\left(x-2011\right)+x\left(x-2011\right)-\left(x-2011\right)+1\)
\(=\left(x^3-x^2+x-1\right)\left(x-2011\right)+1\)
\(\Rightarrow H\left(2011\right)=1\)
Vậy...
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\(\dfrac{x-1}{2012}+\dfrac{x-2}{2013}+\dfrac{x-3}{2014}=\dfrac{x-4}{2015}+\dfrac{x-5}{2016}+\dfrac{x-6}{2017}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2012}+1\right)+\left(\dfrac{x-2}{2013}+1\right)+\left(\dfrac{x-3}{2014}+1\right)=\left(\dfrac{x-4}{2015}+1\right)+\left(\dfrac{x-5}{2016}+1\right)+\left(\dfrac{x-6}{2017}+1\right)\)
\(\Leftrightarrow\dfrac{x+2011}{2012}+\dfrac{x+2011}{2013}+\dfrac{x+2011}{2014}-\dfrac{x+2011}{2015}-\dfrac{x+2011}{2016}-\dfrac{x+2011}{2017}=0\)
\(\Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow x=-2011\)( do \(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\))
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\(\dfrac{x-1}{2014}+\dfrac{x-2}{2013}=\dfrac{x-3}{2012}+\dfrac{x-4}{2011}\)
\(\Leftrightarrow\text{}\text{}\text{}\dfrac{x-1}{2014}-1+\dfrac{x-2}{2013}-1=\dfrac{x-3}{2012}-1+\dfrac{x-4}{2011}-1\)
\(\Leftrightarrow\dfrac{x-2015}{2014}+\dfrac{x-2015}{2013}-\dfrac{x-2015}{2012}-\dfrac{x-2015}{2011}=0\)
\(\Leftrightarrow\left(x-2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\right)=0\)
mà \(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\ne0\)
nên \(x-2015=0\)
\(\Leftrightarrow x=2015\)
x+2.x+3.x+4.x+....+2012.x=2012.2013
x(1+2+3+4+...+2012)=2012.2013
x(2012.2013:2)=2012.2013
x=2012.2013:(2012.2013:2)
x=2
x+2.x+3.x+4.x+....+2012.x=2012.2013
x.(1+2+3+4+...+2012)=2012.2013
Số số hạng từ 1 đến 2012 là:
(2012-1):1+1=2012(số hạng)
Tổng từ 1 đến 2012 là:
(2012+1)x2012:2=2025078
x.(1+2+3+4+...+2012)=2012.2013
x.2025078=4050156
x=4050156:2025078
x=2
Vậy x=2