So sánh
a) \(\frac{53}{57}\)và \(\frac{531}{571}\)
b) \(\frac{2018.2019-1}{2018.2019}\) và \(\frac{2017.2018-1}{2017.2018}\)
So sánh
\(\frac{2017.2018+1}{2017.2018}\) và \(\frac{2018.2019+1}{2018.2019}\)
\(\frac{2017.2018}{2017.2018-1}\) và \(\frac{2018.2019}{2018.2019-1}\)
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 : \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\) => \(\frac{2017.2018+1}{2017.2018}>\frac{2018.2019+1}{2018.2019}\)
a) So sánh M và N:
\(M=\frac{2018}{2019}+\frac{2019}{2020}\)
\(N=\frac{2018+2019}{2019+2020}\)
b) So sánh A và B:
\(A=\frac{2017.2018-1}{2017.2018}\)
\(B=\frac{2018.2019-1}{2018.2019}\)
c) \(\frac{19}{31}\) và \(\frac{17}{35}\)
d) \(\frac{3535}{3534}\) và \(\frac{2323}{2322}\)
Làm nhanh mình đang cần gấp, sáng mai mình đi học thêm !! T.T
Đúng mình sẽ tick
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 !
So sánh : 2017.2018 - 1 / 2017.2018 và 2018.2019 - 1 / 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 \(\frac{2017.2018-1}{2017.2018}< \frac{2018.2019-1}{2018.2019}\)
Tính tổng sau một cách hợp lí :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2017.2018}+\frac{1}{2018.2019}\)
Làm được t tick
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
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2019}{2019}-\frac{1}{2019}\)
\(=\frac{2018}{2019}\)
cái ĐỒ ĐÁNG GHÉT ◥ὦɧ◤ŤŔầŃ VăŃ ĤùŃĞ™ kia t định trả lời sao m dám....
So sánh
C=2017.2018-1/2017.2018 và D=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:
\(\frac{1}{2017.2018}>\frac{1}{2018.2019}\Rightarrow1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\Rightarrow C< D\)
so sanh 2 phan so sau
a=2017.2018-1 / 2017.2018
b=2018.2019-1 / 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=2017.2018−12017.2018a=2017.2018-12017.2018
=1−12017.2018=1-12017.2018
b=2018.2019−12018.2019b=2018.2019-12018.2019
=1−12018.2019=1-12018.2019
Vì 2017.2018<2019.20182017.2018<2019.2018
⇒12017.2018>12019.2018⇒12017.2018>12019.2018
⇒1−12017.2018<1−12019.2018⇒1-12017.2018<1-12019.2018
Hay a<b
a, 2003.2004-1/2003.2004 và.
2004.2005-1/2004.2005
b, 2017.2018/2017.2018+1 và 2018.2019/2018.2019+1
so sánh:\(\frac{53}{57}\)và \(\frac{531}{571}\)
So sánh hai số : \(\frac{53}{57}\)và\(\frac{531}{571}\)
Ta đổi: 53/70 = 530/570
1 - 530/570 = 40/570
1 - 531/571 = 40/571
Vì 40/570 > 4/571
Nên 53/57 < 531/571