\(x.\left(x-2009\right)-2010x+2009.2010=0\)

\(x.\left(x-2009\right)-2010\left(x-2009\right)=0\)

\(\left(x-2009\right)\left(x-2010\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2009=0\\x-2010=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2009\\x=2010\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=2009\\x=2010\end{cases}}\)

Thay 2010 = x + 1 vào P ( x ),ta có :

\(^{x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...+\left(x+1\right)x^2-\left(x+1\right)x-1}\)

= x¹⁰ - x¹⁰ - x⁹ + x⁹ + x⁸ - x⁸ - x⁷ + ... + x³ + x² - x² + x - 1

= x + 1

= 2009 + 1

= 2010
 

Thay 2010 = x+ 1 vào P( x) ,có :

\(x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...+\left(x+1\right)x^2-\left(x+1\right)x-1\)

= \(x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^3+x^2-x^2+x-1\) 

= x+1 

= 2009 + 1

= 2010

Ta có: x = 2011 \(\Rightarrow\) 2010 = x - 1

\(A=x^{2011}-2010x^{2010}-2010x^{2009}-...-2010x+1\)

\(=x^{2011}-\left(x-1\right)x^{2010}-\left(x-1\right)x^{2009}-...-\left(x-1\right)x+1\)

\(=x^{2011}-\left(x-1\right)x^{2010}-\left(x-1\right)x^{2009}-...-\left(x-1\right)x+1\)

\(=x^{2011}-x^{2011}+x^{2010}-x^{2010}+x^{2009}-...-x^2+x+1\)

\(=x+1\)

\(=2011+1\)

\(=2012.\)

x=2011

=> 2010= x-1

A = x^2011- (x-1) x^2010- (x-1).x^2009-.....- (x-1).x+1

= x^2011-x^2011+x^2010- x^2010+x^2009..x^2.-x^2+x+1

= x+1

=(x-1)+2= 2010+2=2012

gì thế

lớp 6 đây

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cưus

k mk nha

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x.\left(x+1\right)}=\frac{2005}{2010}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{401}{402}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{401}{402}\)

\(\Leftrightarrow\frac{1}{x+1}=1-\frac{401}{402}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{402}\)

\(\Leftrightarrow x+1=402\Rightarrow x=401\)