bằng 1

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2009}\)

\(=\frac{1}{\frac{2\cdot\left(1+2\right)}{2}}+\frac{1}{\frac{3\cdot\left(3+1\right)}{2}}+\frac{1}{\frac{4\cdot\left(4+1\right)}{2}}+...+\frac{1}{\frac{2009\cdot\left(2009+1\right)}{2}}\)

\(=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{2009\cdot2010}\)

\(=2\cdot\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2009\cdot2010}\right)\)

\(=2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2010}\right)\)

\(=2\cdot\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(=1-\frac{1}{1005}\)

\(=\frac{1004}{1005}\)

1/1+2=3=1/1+2+2=6=1/1+2+3+4=10+3+6=19+1/1+2+3+4=29+3+6+10+19+2009=2076nếu mình làm sai thì nhớ chỉ dùm

nhớ kết bạn với mình nhé

\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2009}\)

\(=\frac{1}{\left(2+1\right).2:2}+\frac{1}{\left(3+1\right).3:2}+\frac{1}{\left(4+1\right).4:2}+...+\frac{1}{\left(2009+1\right).2009:2}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2009.2010}=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2010}\right)\)\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)=\frac{2.502}{1005}=\frac{1004}{1005}\)

=17/6:(1-2/3)

=17/6:1/3

=17/2

=13/6×9/2-6/7

=39/4-6/7

=249/28

a) \(\left(\frac{5}{2}+\frac{1}{3}\right):\left(1-\frac{2}{3}\right)=\left(\frac{15}{6}+\frac{2}{6}\right):\frac{1}{3}\)

\(=\frac{17}{6}:\frac{1}{3}=\frac{17}{6}\cdot\frac{3}{1}=\frac{17}{2}\cdot\frac{1}{1}=\frac{17}{2}\)

b) \(\left(\frac{5}{2}-\frac{1}{3}\right)\cdot\frac{9}{2}-\frac{6}{7}=\left(\frac{15}{6}-\frac{2}{6}\right)\cdot\frac{9}{2}-\frac{6}{7}\)

\(=\frac{13}{6}\cdot\frac{9}{2}-\frac{6}{7}=\frac{13}{2}\cdot\frac{3}{2}-\frac{6}{7}=\frac{39}{4}-\frac{6}{7}=\frac{273}{28}-\frac{24}{28}=\frac{249}{28}\)

a ) \(\left(\frac{5}{2}+\frac{1}{3}\right)\div\left(1-\frac{2}{3}\right)\)

\(=\left(\frac{15}{6}+\frac{2}{6}\right)\div\frac{1}{3}\)

\(=\frac{17}{6}\div\frac{1}{3}=\frac{17}{6}\cdot\frac{3}{1}\)

\(=\frac{17}{2}\)

b ) \(\left(\frac{5}{2}-\frac{1}{3}\right)\cdot\frac{9}{2}-\frac{6}{7}\)

\(=\left(\frac{15}{6}-\frac{2}{6}\right)\cdot\frac{9}{2}-\frac{6}{7}\)

\(=\frac{13}{6}\cdot\frac{51}{14}\)

\(=\frac{221}{28}\)

\(G=\frac{1}{3^0}+\frac{1}{3^1}+...+\frac{1}{3^{2005}}\)\(\Rightarrow3G=3+\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)

\(\Rightarrow3G-G=2G=3-\frac{1}{3^{2005}}\)\(\Rightarrow G=\frac{3-\frac{1}{3^{2005}}}{2}\)

\(Y=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)\(\Rightarrow2Y=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

\(\Rightarrow2Y-Y=2-\frac{1}{2^{2012}}\) \(\Rightarrow Y=2-\frac{1}{2^{2012}}\)

Cảm ơn bạn nhiều lắm 

Sai đề rùi, mk nghĩ phần cuối mẫu số phải là 98.1

Ờ đúng rồi là 98*1 đó

giải dùm mình với

Ta có:

B=1/2-1/2^2-1/2^3-...-1/2^100

B/2=1/2^2-1/2^3-1/2^4-....-1/2^101

B/2-B=1/2^101-1/2

=>B=(1/2^101-1/2).2

Vậy:B=(1/2^101-1/2).2

Bạn nào làm đc nhanh mình sẽ k

Bạn làm nhanh và đúng nhất mình sẽ k

a) \(D=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)

\(\Rightarrow7D=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\)

\(\Rightarrow7D-D=\left(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)\)

\(\Rightarrow6D=1-\frac{1}{7^{100}}\)

\(\Rightarrow D=\left(1-\frac{1}{7^{100}}\right).\frac{1}{6}\)