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
\(\frac{2017.2018+1}{2017.2018}\) 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 : \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\) => \(\frac{2017.2018+1}{2017.2018}>\frac{2018.2019+1}{2018.2019}\)
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 \(\frac{2017.2018-1}{2017.2018}< \frac{2018.2019-1}{2018.2019}\)
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
So sánh
a) \(\frac{53}{57}\)và \(\frac{531}{571}\)
b) \(\frac{2018.2019-1}{2018.2019}\) và \(\frac{2017.2018-1}{2017.2018}\)
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 \(\frac{53}{57}< \frac{531}{571}\)
Các bạn giúp mk bài này với: (Bài này phải so sánh nhé)
1) 12/49 và 13/47
2) 456/461 và 123/128
3) 64/85 và 73/81
4) 2003.2004/2003.2004 và 2004.2005-1/2004-2005
5) 2016.2017/2016.2017+1 và 2018.2019/2018.2019+1
6) 108+2/108-1 và 108/108-3
7) 244.395-151/244+395.243 và 423134.846267-423133/423133.846267-423134
8) 33.103/23.5.103+7000 và 3774/5217
So sánh: A = \(\dfrac{2003.2004-1}{2003.2004}\) và B = \(\dfrac{2004.2005-1}{2004.2005}\)
Ta có :
+) \(\frac{2003.2004-1}{2003.2004}=\frac{2003.2004}{2003.2004}-\frac{1}{2003.2004}=1-\frac{1}{2003.2004}\)
+) \(\frac{2004.2005-1}{2004.2005}=\frac{2004.2005}{2004.2005}-\frac{1}{2004.2005}=1-\frac{1}{2004.2005}\)
ta thấy :
\(\frac{1}{2003.2004}>\frac{1}{2004.2005}\Rightarrow1-\frac{1}{2003.2004}< 1-\frac{1}{2004.2005}\)
\(\Rightarrow\frac{2003.2004-1}{2003.2004}< \frac{2004.2005-1}{2004.2005}\)
S = 1/1.2 + 1/2.3 + 1/3.4 +..........1/2017.2018 + 1/2018.2019
can gap
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}+\frac{1}{2018.2019}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\)
( 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
S=1-1/2+1/2-1/3+1/3-1/4+.........+1/2017-1/2018+1/2018-1/2019
S=1-1/2019
S=2019/2019-1/2019
S=2018/2019
So sánh 2003.2004-1/2003.2004 và 2004.2005-1/2004.2005
ta có :
+) \(\frac{2003.2004-1}{2003.2004}=\frac{2003.2004}{2003.2004}-\frac{1}{2003.2004}=1-\frac{1}{2003.2004}\)
+) \(\frac{2004.2005-1}{2004.2005}=\frac{2004.2005}{2004.2005}-\frac{1}{2004.2005}=1-\frac{1}{2004.2005}\)
ta thấy :
\(\frac{1}{2003.2004}>\frac{1}{2004.2005}\Rightarrow1-\frac{1}{2003.2004}< 1-\frac{1}{2004.2005}\)
\(\Rightarrow\frac{2003.2004-1}{2003.2004}< \frac{2004.2005-1}{2004.2005}\)