Gọi \(S=\frac{2009}{1}+\frac{2008}{2}+...+\frac{1}{2009}\)

\(\Rightarrow S=\frac{2010-1}{1}+\frac{2010-2}{2}+...+\frac{2010-2009}{2009}\)

\(\Rightarrow S=2010-1+\frac{2010}{2}-1+...+\frac{2010}{2009}-1\)

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

\(\Rightarrow S=2010+\frac{2010}{2}+...+\frac{2010}{2009}-2009\)

\(\Rightarrow S=\frac{2010}{2}+\frac{2010}{3}+...+\frac{2010}{2009}+1\)

\(\Rightarrow S=\frac{2010}{2}+\frac{2010}{3}+..+\frac{2010}{2009}+\frac{2010}{2010}\)

\(\Rightarrow S=2010\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)\)

Khi đó \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}}{2010\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)}=\frac{1}{2010}\)

có : Q = [ 2 + 2^2 ] + [ 2^3 +2^4] + ... + [2^9 +  2^10]

Q = 2 [1+2] +2^3[1 +2]+ ...+ 2^9 [1+2]

Q = 2 . 3+2^3 .3 +... + 2^9 .3

Q = 3. [ 2 + 2^3 +... + 2^9]

Vậy Q chia hết cho 3

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

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

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

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

Câu a:

Có dạng tổng quát:\(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{x+1}}=\frac{1}{\sqrt{\left(k+1\right)k}\left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{\left(k+1\right)k}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k-1}}\)

Áp dụng kết quả trên suy ra câu a

khó wá câu này i don't know

ngu thì đừng bày đặt

Đặt \(A=1+2+2^2+2^3+....+2^{2008}\)

\(2A=2+2^2+2^3+2^4+....+2^{2019}\)

\(A=2^{2019}-1\)

\(\Rightarrow B=\frac{2^{2019}-1}{1-2^{2019}}=\frac{-\left(1-2^{2019}\right)}{1-2^{2019}}=-1\)