\(\Leftrightarrow\dfrac{x+1}{2010}+1+\dfrac{x+2}{2009}+1+...+\dfrac{x+2009}{2}+1+\dfrac{x+2010}{1}+1=0\)

=>x+2011=0

hay x=-2011

Mik nghĩ là C

Chúc bạn hok tốt

Chọn D

D

\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)

\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)

\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)

\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)

\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)

x=-2011

cách bạn giải bài này là gì vậy ?

(2008 x 2009 x 2010 x 2011) x (1 + 1/2 : 3/2 - 4/3)

=(2008 x 2009 x 2010 x 2011) x (1 + 1/3 - 4/3)

=(2008 x 2009 x 2010 x 2011) x (4/3 - 4/3)

=(2008 x 2009 x 2010 x 2011) x 0

=0

đáp số:0

ủng hộ nha

kết ban đi

đáp số: 0

dễ như ăn cháo

thảo mai ăn hủ tiếu mì nha

tìm x hử

Bài 2:

Vì a,b là nghiệm PT nên \(\left\{{}\begin{matrix}30a^2-4a=2010\\30b^2-4b=2010\end{matrix}\right.\)

\(\Rightarrow N=\dfrac{a^{2008}\left(30a^2-4a\right)+b^{2008}\left(30b^2-4b\right)}{a^{2008}+b^{2008}}\\ \Rightarrow N=\dfrac{a^{2008}\cdot2010+b^{2008}\cdot2010}{a^{2008}+b^{2008}}=2010\)

Bài 1:

Viét: \(\left\{{}\begin{matrix}x_1+x_2=a\\x_1x_2=a-1\end{matrix}\right.\)

\(M=\dfrac{2x_1^2+x_1x_2+2x_2^2}{x_1^2x_2+x_1x_2^2}=\dfrac{2\left(x_1+x_2\right)^2-3x_1x_2}{x_1x_2\left(x_1+x_2\right)}=\dfrac{2a^2-3a+3}{a^2-a}\)

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

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

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

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

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

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

Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)

=> x - 2012 = 0

=> x = 2012

Vậy x = 2012

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

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

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

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

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

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

Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)

=> x - 2012 = 0

=> x = 2012

Vậy x = 2012

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

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

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

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

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

\(mà\frac{1}{2011}<\frac{1}{2010}<\frac{1}{2009}<\frac{1}{2008}\Rightarrow\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)

=>x-2012=0

=>x=2012

vậy x=2012