đặt A=3²/20x23+3²/23x26+...+3²/77x80

\(A=3\left(\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+...+\frac{3}{77\cdot80}\right)\)

\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)

\(=3\left(\frac{1}{20}-\frac{1}{80}\right)\)

\(=3\cdot\frac{3}{80}\)

\(=\frac{9}{80}\)

giúp với

da la dau

A=1/20*23+1/23*26+...+1/77*80

=1/3(1/20-1/23+1/23-1/26+...+1/77-1/80)

=1/3*3/80=1/80<1/79

đặt A=9/20x23+9/23x26+...+9/77x80

<=>\(A=3\left(\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+...+\frac{3}{77\cdot80}\right)\)

\(\Rightarrow A=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)

\(\Rightarrow A=3\left(\frac{1}{20}-\frac{1}{80}\right)\)

\(\Rightarrow A=3\cdot\frac{3}{80}\)

\(\Rightarrow A=\frac{9}{80}\)

a) \(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{2007x2009}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2007}-\frac{1}{2009}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2009}\right)=\frac{1}{2}\cdot\frac{2008}{2009}=\frac{1004}{2009}\)

....

các bài cn lại bn lm tương tự nha

b, \(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)

3A = \(\dfrac{1}{6}+\dfrac{1}{18}+...+\dfrac{1}{330}\)

3A-A = \(\dfrac{1}{6}-\dfrac{1}{990}\)

2A = 82/495

A =82/495 : 2 

A=41/495

c, \(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)

A= \(\dfrac{4}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2008.2010}\right)\)

A= \(\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

A= \(\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{1010}\right)\)

A= \(\dfrac{4}{2}.\dfrac{252}{505}\)

A= \(\dfrac{504}{505}\)

\(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+...+\frac{1}{77\cdot80}\)

\(=\frac{1}{3}\left[\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+...+\frac{3}{77\cdot80}\right]\)

\(=\frac{1}{3}\left[\frac{1}{20}-\frac{1}{23}+...+\frac{1}{77}-\frac{1}{80}\right]\)

\(=\frac{1}{3}\left[\frac{1}{20}-\frac{1}{80}\right]\)

\(=\frac{1}{3}\left[\frac{4}{80}-\frac{1}{80}\right]\)

\(=\frac{1}{3}\cdot\frac{3}{80}=\frac{1}{1}\cdot\frac{1}{80}=\frac{1}{80}\)

Mà \(\frac{1}{80}< \frac{1}{9}\)nên \(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+...+\frac{1}{77\cdot80}< \frac{1}{9}\)

Vậy : ...

\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\)

\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)

\(=\frac{1}{3}.\frac{3}{80}\)

\(=\frac{1}{80}< \frac{1}{9}\)

\(\frac{1}{11×14}+\frac{1}{14×17}+\frac{1}{17×20}+\frac{1}{20×23}+\frac{1}{23×26}\)

\(=\frac{1}{3}×\left(\frac{3}{11×14}+\frac{3}{14×17}+\frac{3}{17×20}+\frac{3}{20×23}+\frac{3}{23×26}\right)\)

\(=\frac{1}{3}×\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{23}-\frac{1}{26}\right)\)

\(=\frac{1}{3}×\left(\frac{1}{11}-\frac{1}{26}\right)\)

\(=\frac{1}{3}×\frac{15}{286}\)

\(=\frac{5}{286}\)

\(\frac{1}{11\times14}+\frac{1}{14\times17}+\frac{1}{17\times20}+\frac{1}{20\times23}+\frac{1}{23\times26}\)

\(=\frac{1}{3}\times\left(\frac{1}{11\times14}+\frac{1}{14\times17}+\frac{1}{17\times20}+\frac{1}{20\times23}+\frac{1}{23\times26}\right)\)

\(=\frac{1}{3}\times\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}\right)\)

\(=\frac{1}{3}\times\left(\frac{1}{11}-\frac{1}{26}\right)\)

  =    5/286

Đặt biểu thức trên là A. Ta có :

3A = \(\frac{3}{11x14}+\frac{3}{14x17}+\frac{3}{17x20}+\frac{3}{20x23}+\frac{3}{23x26}\)

3A = \(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+\frac{1}{23}-\frac{1}{26}\)

3A = \(\frac{1}{11}-\frac{1}{26}=\frac{15}{286}\)

A = \(\frac{15}{286}:3=\frac{5}{286}\)

                                  Vậy A = \(\frac{15}{286}\)

ai tả lời giúp mình với mình đang cần gấp

Trong 3 số `2n+1, 2n+2, 2n+3` luôn có một số chia hết cho 3

\(\Rightarrow\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)⋮3\) (1)

Xét \(n⋮2\)

Có: \(2n⋮2,2⋮2\Rightarrow2n+2⋮2\)

\(\Rightarrow\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)⋮2\) (2)

Xét \(n⋮̸2\)

Có: \(2n⋮2\left(dư1\right),1⋮2\left(dư1\right)\Rightarrow2n+1⋮2\)

\(\Rightarrow\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)⋮2\) (3)

Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrowđpcm\)

\(S=1+2+2^2+...+2^{99}\)

\(S=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{98}+2^{99}\right)\)

\(S=3+2^2.3+...+2^{98}.3\)

\(=3\left(1+2^2+...+2^{98}\right)⋮3\)