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`A=1/(299.297)-1/(297.295)-1/(295.293)-...-1/(3.1)`
`2A=2/(299.297)-(2/(297.295)+2/(295.293)+....+2/(3.1))`
`2A=2/(299.297)-(1/295-1/297+1/293-1/295+....+1-1/3)`
`2A=2/(299.297)-(1-1/297)`
`2A=2/(299.297)-296/297`
`A=1/(299.297)-148/297=(1-148.299)/(299.297)`
=> -A = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}-\frac{1}{97.99}\)
=> -2A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}-\frac{2}{97.99}\)
=> \(-2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}-\frac{1}{97}+\frac{1}{99}\)
=> \(-2A=1-\frac{1}{97}-\frac{1}{97}+\frac{1}{99}=\frac{9502}{9603}\)
=> \(A=\frac{9502}{9603}:\left(-2\right)=-\frac{4751}{9603}\)
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}...-\frac{1}{6}-\frac{1}{2}\)
= \(\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+...+\frac{1}{6}+\frac{1}{2}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\frac{8}{9}\)
= \(0\)
"girl cute" là sai rồi bạn ơi, trong tiếng anh, tính từ (cute) phải đứng trước danh tư (girl).
\(M=\frac{1}{9.10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\right)=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)
\(\left(-1-\frac{1}{12}\right).\left(-1-\frac{1}{13}\right).\left(-1-\frac{1}{14}\right)...\left(-1-\frac{1}{2017}\right)\)
\(=\frac{-13}{12}.\frac{-14}{13}.\frac{-15}{14}...\frac{-2018}{2017}\)
\(=\frac{-13}{12}.\frac{14}{-13}.\frac{-15}{14}...\frac{2018}{-2017}\)
\(=\frac{\left(-13\right).14.\left(-15\right)...2018}{12.\left(-13\right).14...2017}=\frac{2018}{12}=\frac{1009}{6}\)
\(\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}+\frac{1}{6}.\frac{1}{7}+\frac{1}{7}.\frac{1}{8}+\frac{1}{8}.\frac{1}{9}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}\)
\(=\frac{9}{18}-\frac{2}{18}\)
\(=\frac{7}{18}\)
\(\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}+\frac{1}{6}.\frac{1}{7}+\frac{1}{7}.\frac{1}{8}+\frac{1}{8}.\frac{1}{9}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}\)
\(=\frac{7}{18}\)
Chúc bạn học tốt !!!!