\(a,x\in\left\{-2021;-2020;...;2020;2021\right\}\\ \Rightarrow\Sigma x=\left(-2021+2021\right)+\left(-2020+2020\right)+....+\left(-1+1\right)+0=0+0+...+0=0\\ b,x\in\left\{-2011;-2010;....;2009;2010\right\}\\ \Rightarrow\Sigma x=-2011+\left(-2010+2010\right)+...+\left(-1+1\right)+0=-2011+0+0+...+0+0=-2011\)

Chọn A

a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)

\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)

\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)

\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\)    (1)

Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)

Nên biểu thức (1) xảy ra khi \(x+2013=0\)

\(x=-2013\)

b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)  (2)

Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)

Nên biểu thức (2) xảy ra khi \(x-2011=0\)

\(x=2011\)

a, 2010 x 3+ 2010 x 6 + 2010

= 2010 x ( 3 + 6 + 1)

= 2010 x 10

= 20100

b, 2011 x 89 + 10 x 2011 + 2011

= 2011 x (89 + 10 + 1)

= 2011 x 100

= 201100

a, 2011 x 3+ 2011 x 6 + 2011

= 2011 x ( 3+6+1)

= 2011 x 10

= 20110

b, 2010 x 89 + 10 x 2010 + 2010

= 2010 x (89+10+1)

= 2010 x 100

= 201000

x¹ + x² + x³ +.....+ x²⁰¹¹ = 0

=> (x¹ + x²) + (x³ + x⁴) +....+(x²⁰⁰⁹ + x²⁰¹⁰) + x²⁰¹¹ = 0

=> 2 + 2 + 2 +.....+ 2 + x²⁰¹¹ = 0

=> 1005 . 2 + x²⁰¹¹ = 0

=> 2010 + x²⁰¹¹ = 0

=> x²⁰¹¹ = -2010

=> Không có giá trị nào của x thỏa mãn đề bài

x¹ + x² + x³ +.....+ x²⁰¹¹

= (x¹ + x²) + (x³ + x⁴) +....+ ( x²⁰⁰⁹ + x²⁰¹⁰) + x²⁰¹¹

= 2 + 2 + 2+.....+ 2 + x²⁰¹¹

= 1005 . 2 + x²⁰¹¹

=2010 + x²⁰¹¹ = 0

=> x²⁰¹¹ = -2010

=> Không có giá trị của x thỏa mãn đề bài