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\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3.\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{94.97}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(=1-\dfrac{1}{97}\)
\(=\dfrac{96}{97}\)
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
\(=3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\right)\)
\(=3\left(1-\dfrac{1}{97}\right)\)
\(=3.\dfrac{96}{97}=\dfrac{288}{97}\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{43.46}\\ S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\\ S=1-\dfrac{1}{46}< 1\)
Vậy S < 1 (đpcm)
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}\\ =1-\dfrac{1}{43}\\ =\dfrac{42}{43}\)
\(=-\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{61}-\dfrac{1}{64}\right)=-\dfrac{1}{63}\)
\(S=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{43\cdot46}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\)
\(S=1-\dfrac{1}{46}< 1\)
S= \(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{40\cdot43}+\dfrac{3}{43\cdot46}\)
S= \(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{42}-\dfrac{1}{46}\)
S= \(1-\dfrac{1}{46}\)
S= \(\dfrac{45}{46}\)
Mà \(\dfrac{45}{46}< 1\)
\(\Rightarrow S< 1\)
Vậy S < 1
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=\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{97.100}\)
=\(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(B=1-\dfrac{3}{1\cdot4}-\dfrac{3}{4\cdot7}-...-\dfrac{3}{2020\cdot2023}\\ =1-\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2020\cdot2023}\right)\\ =1-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2020}-\dfrac{1}{2023}\right)\\ =1-\left(1-\dfrac{1}{2023}\right)\\ =1-\dfrac{2022}{2023}=\dfrac{1}{2023}\)
`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`
\(\dfrac{3}{1\times4}x+\dfrac{3}{4\times7}x+\dfrac{3}{7\times10}x+...+\dfrac{3}{31\times34}x=33\)
\(x\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+...+\dfrac{3}{31\times34}\right)=33\)
\(x\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=33\)
\(x\left(1-\dfrac{1}{34}\right)=33\)
\(\dfrac{33}{34}x=33\)
\(x=34\)
\(\dfrac{3}{1.4}x+\dfrac{3}{4.7}x+\dfrac{3}{7.10}x+...+\dfrac{3}{31.34}x=33\)
\(x.3\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{31.34}\right)=33\)
\(x.3.\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=33\)
\(x.\left(1-\dfrac{1}{34}\right)=33\)
\(x.\dfrac{33}{34}=33\)
\(x=33:\dfrac{33}{34}=33.\dfrac{34}{33}\)
\(x=34\)
\(S=\) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{97.100}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+....+\dfrac{1}{97}-\dfrac{1}{100}\)
(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!!!