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

=>x+2011=0

hay x=-2011

\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)

\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)

\(=\frac{1}{5}+\frac{2}{3}\)

\(=\frac{13}{15}\)

Câu 1:

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)..\left(1-\frac{1}{2016}\right)\left(1-\frac{1}{2017}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2015}{2016}.\frac{2016}{2017}\)

\(A=\frac{1}{2017}\)

Vậy ..............................

Phần giống nhau là gạch ý!

Câu 2

\(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\)

\(\Rightarrow S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)

Đặt \(Q=2^{2009}+2^{2008}+...+2+1\)

\(\Rightarrow2Q=2^{2010}+2^{2009}+...+4+2\)

\(\Rightarrow2Q-Q=\left(2^{2010}+2^{2009}+...+4+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)

\(\Rightarrow Q=2^{2010}-1\)

\(\Rightarrow S=2^{2010}-\left(2^{2010}-1\right)\)

\(\Rightarrow S=2^{2010}-2^{2010}+1\)

\(\Rightarrow S=1\)

Vậy .........................

b) \(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\)

\(\Rightarrow2S=2^{2011}-2^{2010}-2^{2009}-...-2^2-2\)

\(\Rightarrow2S-S=\left(2^{2011}-2^{2010}-2^{2009}-...-2^2-2\right)-\left(2^{2010}-2^{2009}-2^{2008}-...-2-1\right)\)

\(\Rightarrow S=2^{2011}-2^{2010}-2^{2009}-...-2^2-2-2^{2010}+2^{2009}+2^{2008}+...+2+1\)

\(\Rightarrow S=2^{2011}-2^{2010}-2^{2010}+1\)

\(\Rightarrow S=2^{2011}-2.2^{2010}+1\)

\(\Rightarrow S=2^{2011}-2^{2011}+1\)

\(\Rightarrow S=0+1\)

\(\Rightarrow S=1.\)

Vậy \(S=1.\)

Chúc bạn học tốt!

a, \(\frac{1}{2009}+\frac{2}{2009}+...+\frac{2008}{2009}\\ \frac{\left(1+2008\right)\cdot2008\div2}{2009}=\frac{2017036}{2009}\)

a,S1=1+(-2)+3+(-4)+..........+2009+(-2010)

S1=-1.(2010:2)

S1=-1005

b,S2=1+(-2)+(-3)+4+5+(-6)+(-7)+............+2008+2009+(-2010)

S2=-1.(2010:2)

S2=-1.1005

S2=-1005