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1.
E = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.10}\) + \(\dfrac{3}{10.13}\) + \(\dfrac{3}{13.16}\) + \(\dfrac{3}{16.19}\) + \(\dfrac{3}{19.22}\)
E = 1 - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... +\(\dfrac{1}{19}\) - \(\dfrac{1}{22}\)
E = 1 - \(\dfrac{1}{22}\)
E = \(\dfrac{21}{22}\)
2.
(x - 4)(x - 5) = 0
TH1:
x - 4 = 0 => x = 4
TH2:
x - 5 = 0 => x = 5
Vậy: x = 4 hoặc x = 5
\(B=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)=\dfrac{5}{3}\left(\dfrac{102}{103}\right)=\dfrac{170}{103}\)
\(\dfrac{3}{2}\)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=\(\dfrac{68}{103}\)
\(B=1-\dfrac{3}{1\cdot4}-\dfrac{3}{4\cdot7}-...-\dfrac{3}{2020\cdot2023}\\ =1-\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2020\cdot2023}\right)\\ =1-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2020}-\dfrac{1}{2023}\right)\\ =1-\left(1-\dfrac{1}{2023}\right)\\ =1-\dfrac{2022}{2023}=\dfrac{1}{2023}\)
`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`
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}\\ =1-\dfrac{1}{43}\\ =\dfrac{42}{43}\)
\(=-\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{61}-\dfrac{1}{64}\right)=-\dfrac{1}{63}\)
`3.11/(1.4) + 11/4.7 + ... + 11/(61.64)`
`= 3.(11/(1.4) + 11/4.7 + ... + 11/(61.64))`
`= 11.(3/(1.4) + 3/4.7 + ... + 3/(61.64))`
`= 11.(1 -1/4 + 1/4 - 1/7 + ... + 1/61 - 1/64)`
`= 11.(1 - 1/64)`
`= 11.63/64`
`= 693/64:3`
`= 231/64`
sửa đề đề viết nhầm.
Sửa đề: 11/1*4+11/4*7+...+11/61*64
=11/3(3/1*4+3/4*7+...+3/61*64)
=11/3(1-1/4+1/4-1/7+...+1/61-1/64)
=11/3*63/64
=11*21/64=231/64
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=\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{97.100}\)
=\(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+.....+\dfrac{5}{100.103}\)
\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(B=\dfrac{5}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(B=\dfrac{5}{3}.\left(1-\dfrac{1}{103}\right)\)
\(B=\dfrac{5}{3}.\dfrac{102}{103}\)
\(B=1\dfrac{67}{103}\)
Ta có: \(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
\(=\dfrac{5}{3}\left(1-\dfrac{1}{4}\right)+\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}\right)+...+\dfrac{5}{3}\left(\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{3}.\dfrac{102}{103}=\dfrac{170}{103}\)
Vậy \(B=\dfrac{170}{103}\).
\(B=\dfrac{4}{1.4}+\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{100.103}+\dfrac{4}{103.106}\)
\(B=4\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}+\dfrac{1}{103.106}\right)\)
\(B=\dfrac{4}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}+\dfrac{3}{103.106}\right)\)
\(B=\dfrac{4}{3}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}+\dfrac{1}{103}-\dfrac{1}{106}\right)\)
\(B=\dfrac{4}{3}\left(\dfrac{1}{3}-\dfrac{1}{106}\right)\)
\(B=\dfrac{4}{3}.\dfrac{103}{318}\)
\(B=\dfrac{412}{954}\)