\(\dfrac{30}{-x}=-\dfrac{6}{5}\)

\(\dfrac{30}{-x}=\dfrac{-6\cdot5}{5\cdot5}\)

\(\dfrac{30}{-x}=\dfrac{-30}{25}\)

\(\Rightarrow\dfrac{30}{-x}=\dfrac{30}{-25}\)

\(\Rightarrow x=25\)

30/(-x) = -6/5

-x.(-6) = 30.5

6x = 150

x = 150 : 6

x = 25

-12/18 - (-21/35)

= -2/3 + 3/5

= -10/15 + 9/15

= -1/15

\(\dfrac{-12}{18}\) - \(\dfrac{-21}{35}\) 

=  \(\dfrac{-2}{3}\) + \(\dfrac{3}{5}\)

= \(\dfrac{-10}{15}\) + \(\dfrac{9}{15}\)

= \(-\dfrac{1}{15}\)

S =  \(\dfrac{1}{1.3}\)+\(\dfrac{1}{2.4}\)+...+\(\dfrac{1}{97.99}\)+\(\dfrac{1}{98.100}\) - \(\dfrac{49}{99}\)

S = (\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{97.99}\))+(\(\dfrac{1}{2.4}\)+\(\dfrac{1}{4.6}\)+\(\dfrac{1}{6.8}\)+...+\(\dfrac{1}{98.100}\))- \(\dfrac{49}{99}\)

S = \(\dfrac{1}{2}\).(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+...+\(\dfrac{2}{97.99}\))+\(\dfrac{1}{2}\)(\(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))-\(\dfrac{49}{99}\)

S =\(\dfrac{1}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+...+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\))+\(\dfrac{1}{2}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\)-\(\dfrac{1}{100}\))-\(\dfrac{49}{99}\)

S = \(\dfrac{1}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{99}\))+\(\dfrac{1}{2}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{100}\)) - \(\dfrac{49}{99}\)

S = \(\dfrac{1}{2}\).\(\dfrac{98}{99}\) + \(\dfrac{1}{2}\).\(\dfrac{49}{100}\) - \(\dfrac{49}{99}\)

S = \(\dfrac{49}{99}\) + \(\dfrac{49}{200}\) - \(\dfrac{49}{99}\)

S = (\(\dfrac{49}{99}\)- \(\dfrac{49}{99}\)) + \(\dfrac{99}{200}\)

S = 0 + \(\dfrac{49}{200}\)

S = \(\dfrac{49}{200}\)

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 $A<B$