Bài \(13\):
\(C=\dfrac{3}{\left(1\cdot2\right)^2}+\dfrac{5}{\left(2\cdot3\right)^2}+\dfrac{7}{\left(3\cdot4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{n^2+2n+1-n^2}{n^2\left(n+1\right)^2}\)
\(=\dfrac{4-1}{1\cdot4}+\dfrac{9-4}{4\cdot9}+\dfrac{16-9}{9\cdot16}+...+\dfrac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)
\(=1-\dfrac{1}{\left(n+1\right)^2}=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)

Bài \(10\):
\(B=\dfrac{5}{2\cdot1}+\dfrac{4}{1\cdot11}+\dfrac{3}{11\cdot2}+\dfrac{1}{2\cdot15}+\dfrac{13}{15\cdot4}\)
\(=7\left(\dfrac{5}{2\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{1}{14\cdot15}+\dfrac{13}{15\cdot28}\right)\)
\(=7\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\right)\)
\(=7\left(\dfrac{1}{2}-\dfrac{1}{28}\right)=7\cdot\dfrac{13}{28}=\dfrac{13}{4}\)

-57.39-75.-21

-648

-57.(75 - 36) - 75.(36 - 57)

= -57.75 + 57.36 - 75.36 + 75.57

= (-57.75 + 75.57) +36.(57 - 75)

= 0 + 36.(-18)

= -648

1) \(\dfrac{6}{7}\cdot\dfrac{8}{13}+\dfrac{6}{13}\cdot\dfrac{9}{7}-\dfrac{3}{13}\cdot\dfrac{6}{7}\)

\(=\dfrac{6}{7}\cdot\dfrac{8}{13}+\dfrac{9}{13}\cdot\dfrac{6}{7}-\dfrac{3}{13}\cdot\dfrac{6}{7}\)

\(=\dfrac{6}{7}\cdot\left(\dfrac{8}{13}+\dfrac{9}{13}-\dfrac{3}{13}\right)\)

\(=\dfrac{6}{7}\cdot\dfrac{14}{13}\)

\(=\dfrac{12}{13}\)

2) \(\dfrac{3}{11}\cdot\dfrac{7}{9}+\dfrac{17}{11}\cdot\dfrac{3}{19}-\dfrac{3}{19}\cdot\dfrac{25}{11}\)

\(=\dfrac{7}{33}+\dfrac{3}{19}\cdot\left(\dfrac{17}{11}-\dfrac{25}{11}\right)\)

\(=\dfrac{7}{33}+\dfrac{3}{19}\cdot\dfrac{-8}{11}\)

\(=\dfrac{7}{33}+\dfrac{-24}{209}\)

\(=\dfrac{61}{627}\) 

3) \(\dfrac{5}{7}\cdot\dfrac{-2}{11}+\dfrac{5}{7}\cdot\dfrac{9}{11}-\dfrac{5}{7}\) 

\(=\dfrac{5}{7}\cdot\left(\dfrac{-2}{11}+\dfrac{9}{11}-1\right)\)

\(=\dfrac{5}{7}\cdot-\dfrac{4}{11}\)

\(=\dfrac{-20}{77}\)

4) \(\dfrac{3}{13}\cdot\dfrac{15}{11}+\dfrac{3}{11}\cdot\dfrac{7}{13}-\dfrac{3}{13}\)

\(=\dfrac{3}{13}\cdot\dfrac{15}{11}+\dfrac{7}{11}\cdot\dfrac{3}{13}-\dfrac{3}{13}\)

\(=\dfrac{3}{13}\cdot\left(\dfrac{15}{11}+\dfrac{7}{11}-1\right)\)

\(=\dfrac{3}{13}\cdot\left(\dfrac{22}{11}-1\right)\)

\(=\dfrac{3}{13}\cdot1\)

\(=\dfrac{3}{13}\)

Lời giải:
Ta có:

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2021^2}+\frac{1}{2022^2}<\underbrace{ \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2021.2022}}_{M}\)

\(M=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{2022-2021}{2021.2022}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2021}-\frac{1}{2022}=1-\frac{1}{2022}<1\)

\(\Rightarrow \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2021^2}+\frac{1}{2022^2}< M< 1\)

Ta có đpcm.

Lời giải:

\(A=\frac{10^{2021}+1}{10^{2020}+1}=\frac{10(10^{2020}+1)-9}{10^{2020}+1}=10-\frac{9}{10^{2020}+1}<10-\frac{9}{10^{2021}+1}=\frac{10^{2022}+1}{10^{2021}+1}=B\)

Vậy $A<B$

-3/7 + 17/26 - (2/13 - 3/7)

= -3/7 + 17/26 - 2/13 + 3/7

= (-3/7 + 3/7) + (17/26 - 2/13)

= 0 + 13/26

= 1/2

bằng 1/2 nha bạn

Cho mình hỏi là n thuộc N hay Z vậy?

soss

=9

9:3(1+2)=3(1+2)=3.3=9

Cái này là 2 đường chéo bằng nhau hay thế nào hả bạn?

bằng nhau nha bạn. mong bạn giúp mình