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\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(A=2\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)
\(A=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=2\left(1-\frac{1}{100}\right)\)
\(A=2.\frac{99}{100}=..............\)
Tự làm nốt nha
a) \(16+\left(27-7\cdot6\right)-\left(94\cdot7-27\cdot99\right)\)
\(=16+27-7\cdot6-94\cdot7+27\cdot99\)
\(=16+27\left(1+99\right)-7\left(6+94\right)=16+2700-700=2016\)
b)\(A=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)
\(=\frac{1}{3}\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-\frac{2}{7}+\frac{2}{7}-\frac{2}{10}+...+\frac{2}{97}-\frac{2}{100}\right)\)
\(=\frac{1}{3}\left(2-\frac{2}{100}\right)=\frac{1}{3}\cdot\frac{99}{50}=\frac{33}{50}\)
\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3.\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)
\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+...+\dfrac{2}{97.100}\)
=> \(\dfrac{2.3}{1.4}+\dfrac{2.3}{4.7}+...+\dfrac{2.3}{97.100}\)
=> \(2.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
=> \(2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
=> \(2.\left(1-\dfrac{1}{100}\right)\)
=>\(2\).\(\dfrac{99}{100}\)
=\(\dfrac{99}{50}\)
\(B=1-\dfrac{3}{1\cdot4}-\dfrac{3}{4\cdot7}-...-\dfrac{3}{2020\cdot2023}\\ =1-\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2020\cdot2023}\right)\\ =1-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2020}-\dfrac{1}{2023}\right)\\ =1-\left(1-\dfrac{1}{2023}\right)\\ =1-\dfrac{2022}{2023}=\dfrac{1}{2023}\)
`B=1-3/(1.4)-3/(4.7)-3/(7.10)-....-3/(2020.2023)`
`B=1-(3/(1.4)+3/(4.7)+.....+3/(2020.2023))`
`B=1-(1-1/4+1/4-1/7+.....+1/2020-1/2023)`
`B=1-(1-1/2023)`
`B=1-1+1/2023=1/2023`
`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`
\(S=\) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{97.100}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+....+\dfrac{1}{97}-\dfrac{1}{100}\)
(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!!!
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=\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{97.100}\)
=\(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}=\dfrac{99}{100}\)