Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 \(\frac{53}{57}< \frac{531}{571}\)
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 !
gọi biểu thức trên là A A=1/1 -1/2+1/3-1/4+...+1/2017-12018+1/2018-1/2019 A=1/1-1/2019 A=2018/2019
1/1.2+1/2.3+1/3.4+1/4.5+...+1/2017.2018+1/2018.2019
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2019}{2019}-\frac{1}{2019}\)
\(=\frac{2018}{2019}\)
\(\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)
\(\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)
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 \(\frac{2017.2018-1}{2017.2018}< \frac{2018.2019-1}{2018.2019}\)
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:
\(\frac{1}{2017.2018}>\frac{1}{2018.2019}\Rightarrow1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\Rightarrow C< D\)
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`
xet bt A ta co
A=2016.2017+1/2016.2017
=1+1/2016.2017
xet bt B ta co
B=2017.2018+1/2017.2018
=1+1/2017.2018
vì 1/2016.2017>1/2017.2018
nen 1+1/2016.2017>1+1/2017.2018
suy ra A>B
ai thay mik lam đúng thì k cho mik nha
555555555555500000000000000.................
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 : \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\) => \(\frac{2017.2018+1}{2017.2018}>\frac{2018.2019+1}{2018.2019}\)