3/5 . 11/7 + 3/5 . (- 4/7) + 2/5
= 3/5 . [11/7 + (- 4/7)]       + 2/5
= 3/5 .        1                     + 2/5
=          3/5                        + 2/5
=                          1

i giúp em vớiiiiii

\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)

\(=\dfrac{3}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{3}{\dfrac{3\left(3+1\right)}{2}}+...+\dfrac{3}{\dfrac{2022\left(2022+1\right)}{2}}\)

\(=\dfrac{6}{2\left(2+1\right)}+\dfrac{6}{3\left(3+1\right)}+...+\dfrac{6}{2022\cdot2023}\)

\(=\dfrac{6}{2\cdot3}+\dfrac{6}{3\cdot4}+...+\dfrac{6}{2022\cdot2023}\)

\(=6\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\)

\(=6\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)=6\cdot\dfrac{2021}{4046}=\dfrac{12126}{4046}< 3\)

mà \(3< \dfrac{10}{3}\)

nên \(M< \dfrac{10}{3}\)

mk làm phần a thui nhé

a. A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6

A = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6

A =  1/2 - 1/6

A= 3/6 - 1/6

A = 1/3

\(B=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\)

\(b=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)

\(b=\frac{1}{2}-\frac{1}{14}\)

\(b=\frac{3}{7}\)

\(d=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)

\(d=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}\)

\(d=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

\(d=1-\frac{1}{11}\)

\(d=\frac{10}{11}\)

\(e=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(e=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+\frac{1}{17\cdot20}\)

\(e=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{17\cdot20}\right)\)

\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(e=\frac{1}{3}\cdot\frac{9}{20}=\frac{3}{20}\)

Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)

\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{90}\)

\(=1+\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{9.10}\right)\)

\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)

\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=1+2\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=1+1-\frac{1}{5}\)(nhân phá ngoặc)

\(=2-\frac{1}{5}\)< 2 

Vậy K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)< 2

-3 > -5

4 > -6

10 > -10

3 < 5

-3 > -5

4> -6

10 > -10

3<5

-3 > -5

4 > -6

10 > -10

3 < 5