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\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+...+\dfrac{2}{97.100}\)
=> \(\dfrac{2.3}{1.4}+\dfrac{2.3}{4.7}+...+\dfrac{2.3}{97.100}\)
=> \(2.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
=> \(2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
=> \(2.\left(1-\dfrac{1}{100}\right)\)
=>\(2\).\(\dfrac{99}{100}\)
=\(\dfrac{99}{50}\)
1.
`16 + (27 - 7.6 ) - (94 -7 - 27.99)`
`= 16+ 27 - 7.6 - 94 + 7 + 27.99`
`= 16 + 27(99 +1) - 7(6-1) - 94`
`= -78 + 27.100 - 7.5`
`= 2587`
2.
`A = 2/1.4 + 2/4.7 + 2/7.10 +...+ 2/97.100`
`A= 2(1/1.4 + 1/4.7 + 1/7.10 +...+1/97.100)`
`3A = 2 (3/1.4 + 3/4.7 + 3/7.10+...+ 3/97.100)`
`3/2 A = 1 - 1/4 + 1/4 - 1/7 +...+ 1/97 - 1/100`
`3/2A = 1 - 1/100`
`3/2 A= 99/100`
`A= 99/100 : 3/2`
`A=33/50`
Vậy `A= 33/50`
1.16+(27-7.6)-(94-7-27.99)=16+27-7.6-94+7+27.99
=(27+27.99)+(27+7-94)+16
=27.100-60+16
=2700-44=2656
2.A=\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{97.100}\)
=\(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
`S_1 = 5/(1.4) + 5/(4.7) +...+ 5/(97.100)`
`S_1 = 5 (1/(1.4) + 1/(4.7) +...+ 1/(97.100))`
`S_1 = 5/3 (3/(1.4) + 3/(4.7) +...+ 3/(97.100))`
`S_1 = 5/3 (1 - 1/4 + 1/4 - 1/7 + ...+ 1/97 - 1/100)`
`S_1 = 5/3 (1 - 1/100)`
`S_1 = 5/3 . 99/100`
`S_1 = 33/20`
\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3.\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)
\(S=\) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{97.100}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+....+\dfrac{1}{97}-\dfrac{1}{100}\)
(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với mọi \(a\in N\)*)
\(S=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Vậy \(S=\dfrac{99}{100}\)
Chúc bạn học tốt!!!
\(a,\dfrac{-8}{15}+\dfrac{13}{30}-\dfrac{5}{12}=\dfrac{-32}{60}+\dfrac{26}{60}-\dfrac{25}{60}=-\dfrac{31}{60}\\ b,\dfrac{3}{2}.\dfrac{7}{2}+\left(\dfrac{-5}{6}+\dfrac{1}{10}:\dfrac{11}{30}\right)=\dfrac{21}{4}+\left(\dfrac{-5}{6}+\dfrac{3}{11}\right)=\dfrac{21}{4}+\dfrac{-37}{66}=\dfrac{619}{132}\)
\(c,\dfrac{-20}{21}.\dfrac{22}{35}+\dfrac{-20}{21}.\dfrac{13}{35}+\dfrac{-22}{21}=\dfrac{-20}{21}\left(\dfrac{22}{35}+\dfrac{13}{35}\right)+\dfrac{-22}{21}=\dfrac{-20}{21}.1+\dfrac{-22}{21}=\dfrac{-20}{21}+\dfrac{-22}{21}=\dfrac{-42}{21}=-2\)
\(\dfrac{3}{2}\)B= \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(\dfrac{3}{2}\)B= \(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\) \(\dfrac{3}{2}\)B= \(\dfrac{102}{103}\) \(\)B= \(\dfrac{102}{103}:\dfrac{3}{2}\) B=\(\dfrac{68}{103}\)
Bài 2 :
a, \(x=\dfrac{3}{5}-\dfrac{7}{8}=\dfrac{24-30}{40}=-\dfrac{6}{40}=-\dfrac{3}{20}\)
b, \(2x-1=-2\Leftrightarrow x=-\dfrac{1}{2}\)
=\(2.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+.....+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
=\(2.\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\)
= \(2.\dfrac{99}{100}\)
=\(\dfrac{99}{50}\)