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Ta có : \(\frac{2017.2018+1}{2017.2018}=1+\frac{1}{2017.2018}\)
\(\frac{2018.2019+1}{2018.2019}=1+\frac{1}{2018.2019}\)
Mà : \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\) => \(\frac{2017.2018+1}{2017.2018}>\frac{2018.2019+1}{2018.2019}\)
Ta có:
\(C=\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)
\(D=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)
Mà ta có:
\(\frac{1}{2017.2018}>\frac{1}{2018.2019}\Rightarrow1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\Rightarrow C< D\)
\(\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)
\(\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)
Ta thấy \(2017.2018< 2018.2019\)
nên \(\frac{1}{2017.1018}>\frac{1}{2018.2019}\)
\(\Rightarrow\)\(1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)
Vậy \(\frac{2017.2018-1}{2017.2018}< \frac{2018.2019-1}{2018.2019}\)
a) \(\frac{53}{57}=\frac{530}{570}\)
Ta có : 1 - \(\frac{530}{570}\)= \(\frac{40}{570}\) ; 1 - \(\frac{531}{571}=\frac{40}{571}\)
Vì \(\frac{40}{570}>\frac{40}{571}\) nên \(\frac{53}{57}< \frac{531}{571}\)
(1.2 + 2.3 + 3.4 + ... + 2018.2019) - (12 + 22 + ... + 20182)
= (1.2 + 2.3 + ... + 2018.2019) - (1.1 + 2.2 + ... + 2018.2018)
= (1.2 + 2.3 + ... + 2018.2019) - [1.(2 - 1) + 2.(3 - 1) + ... + 2018.(2019 - 1)]
= (1.2 + 2.3 + ... + 2018.2019) - (1.2 + 2.3 + ... + 2018.2019 - 1 - 2 - 3 - ... - 2018)
= (1.2 + 2.3 + ... + 2018.2019) - [1.2 + 2.3 + ... + 2018.2019 - (1 + 2 + ... + 2018)]
= (1.2 + 2.3 + ... + 2018.2019) - (1.2 + 2.3 + ... + 2018.2019) + (1 + 2 + 3 + ... + 2018)
= 1 + 2 + ... + 2018 (có : (2018 - 1) : 1 + 1 = 2018 (số))
= (2018 + 1).2018 : 2
= 2037171
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}+\frac{1}{2018.2019}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\)
( gạch bỏ các phân số giống nhau)
\(S=1-\frac{1}{2019}\)
\(S=\frac{2018}{2019}\)
CHÚC BN HỌC TỐT!!!!
S=1/1.2+1/2.3+1/3.4+............1/2017.2018+1/2018.2019
S=1/2.(1+1/3.2+1/3.2+.............1/2017.1009+1/1009.2019)
S=1/4.(2+2/3.2+2/3.2+..............2/2017.1009+2/1009.2019)
S=1/4.(1-1/2+1/2-1/3+1/3+..........+1/1009-1/1009+1/2019)
S=1/4.(1-1/2019)
S=1/4.2018/2019=1009/4038
a) Ta có :
N = 2018 + 2019/2019 + 2020
= 2018/2019 + 2020 + 2019/2019 + 2020
Ta thấy : 2018/2019 + 2020 < 2018/2019 ( Vì 2019 + 2020 > 2019 )
2019/2019 + 2020 < 2019/2020 ( Vì 2019 + 2020 > 2020 )
=> 2018/2019 + 2020 + 2019/2019 + 2020 < 2018/2019 + 2019/2020
=> M > N
b) Mk ko bt làm !!
c) Ta có :
19/31 > 1/2
17/35 < 1/2
=> 19/31 > 17/35
d) Ta có :
3535/3434 = 1 + 1/3534
2323/2322 = 1 + 1/2322
Ta thấy :
1/3534 < 1/2322 ( Vì 3534 > 2322 )
=> 1 + 1/3534 < 1 + 1/2322
=> 3535/3534 < 2323/2322
Hok tốt !
Tính: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}+\dfrac{1}{2018.2019}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}+\dfrac{1}{2018.2019}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\\ =1-\dfrac{1}{2019}\\ =\dfrac{2019-1}{2019}=\dfrac{2018}{2019}\)
`a=(2017.2018-1)/(2017.2018)`
`=1-1/(2017.2018)`
`b=(2018.2019-1)/(2018.2019)`
`=1-1/(2018.2019)`
Vì `2017.2018<2019.2018`
`=>1/(2017.2018)>1/(2019.2018)`
`=>1-1/(2017.2018)<1-1/(2019.2018)`
Hay `a<b`
a<b vì (0,9999997543<0,9999997546)