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\(M=\dfrac{15}{7\cdot8}-\dfrac{17}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}\)
\(M=\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)
\(M=\dfrac{1}{7}-\dfrac{1}{11}=\dfrac{11-7}{7\cdot11}=\dfrac{4}{77}\)
`M=15/(7.8)-17/(8.9)+1/(9.10)+1/(10.11)`
`=1/7-1/8-(1/8-1/9)+1/9-1/10+1/10-1/11`
`=1/7-1/4+2/9-1/11`
`=67/2772`
Theo mình đề bài sai phải là `M=15/(7.8)-17/(8.9)+1/(9.10)+1/(10.11)`
`=1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11`
`=1/7-1/11=4/77`


\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{13\cdot14}+\frac{1}{14\cdot15}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}\)
\(=\frac{1}{2}-\frac{1}{15}\)
\(=\frac{13}{30}\)

Ta có: A = 1.2 + 2.3 + 3.4 + 4.5 +....+ 98.99
⇒⇒ 3A = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 +....+ 98.99.3
⇒⇒ 3A = 1.2.3 + 2.3(4-1) + 3.4(5-2) + 4.5(6-3) +.....+ 98.99(100-97)
⇒⇒ 3A = 1.2.3 + 2.3.4 - 2.3.1 + 3.4.5 - 3.4.2 + 4.5.6 - 4.5.3 + ....+ 98.99.100 - 98.99.97
⇒⇒ 3A = 98.99.100
⇒⇒ A = 98.99.100398.99.1003 = 323400
b, tự giải nhé

a, A = 1.2 + 2.3 + 3.4 + 4.5 + 5.6 + 6.7 + 7.8 + 8.9 + 9.10 + 10.11
= (1.2 + 2.3) + (3.4 + 4.5) + (5.6 + 6.7) + (7.8 + 8.9) + (9.10 + 10.11)
= 2( 1 + 3) + 4( 3 + 5) + 6( 5 + 7) + 8 ( 7 + 9) + 10( 9 + 11)
= 2.4 + 4.8 + 6.12 + 8.16 + 10.20 = 2.2.2 + 2.4.4 + 2.6.6 + 2.8.8 + 2.10.1
b,tương tự nhé

\(=\dfrac{1}{9}-\dfrac{1}{10}-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=\dfrac{1}{9}-\dfrac{1}{10}-\left(1-\dfrac{1}{9}\right)\)
\(=\dfrac{1}{9}-\dfrac{1}{10}-1+\dfrac{1}{9}\)
\(=-\dfrac{97}{90}\)

\(M=\frac{1}{9.10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\right)=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)