Cho 2 phân số:A=\(\frac{201620162016}{201720172017}\) và B=\(\frac{2016}{2017}\)
So sánh A và B.
BÀI 1 : Tính nhanh : \(1\frac{1}{3}\)x \(1\frac{1}{8}\)x \(1\frac{1}{15}\)x ..... ( có 98 thừa số )
BÀI 2 : Tính : \(\frac{1}{24}\)+\(\frac{1}{48}\)+ .... + \(\frac{1}{39600}\)
BÀI 3 : Tính : Cho hai phân số : A = \(\frac{201620162016}{201720172017}\)và B = \(\frac{2016}{2017}\). Hãy so sánh A và B.
GIÚP MÌNH VỚI MÌNH SẼ TICK VÀ KẾT BẠN !! ^-^
Bai3
201620162016/201720172017=2016.100010001/2017.100010002=2016/2017
Vay 201620162016/201720172017=2016/2017
bài 1 kobik
bài 2\(\frac{1}{39600}\):\(\frac{1}{4}\)=\(\frac{2}{33}\)
bài 3\(\frac{201620162016}{201720172017}=\frac{2016}{2017}\)
nên mó bằng nhau
So sánh hai phân số : A=\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)và B=\(\frac{2015+2016+2017}{2016+2017+2018}\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)
\(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Ta có:
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
Cộng vế theo vế, ta có:
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(hay\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Vậy A > B
So sánh hai phân số A=\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)và B=\(\frac{2015+2016+2017}{2016+2017+2018}\)
Các bn giải giúp mk nha ! Mk cần gấp ! Thanks nhiều. ^-^
B = \(\frac{2015+2016+2017}{2016+2017+2018}=\frac{2016.3}{2017.3}=\frac{2016}{2017}\left(1\right)\)
Mà A = \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}.\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)=> A > B.
Vậy A > B .
Bạn Dont look at me
Bạn nên làm theo bạn ấy
Bạn k đúng cho bạn ấy. Bởi vì bạn ấy làm đúng
Theo mk là vậy
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)\(B=\frac{2015+2016+2017}{6051}\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)\(B=\frac{2015}{6051}+\frac{2016}{6051}+\frac{2017}{6051}\)
=> A > B
cho A=\(\frac{-2016}{10^{2016}}+\frac{-2017}{10^{2017}}\)
B=\(\frac{-2017}{10^{2016}}-\frac{2016}{10^{2017}}\)
so sánh A và B
Lấy A - B ta được
\(A-B=\frac{-2016}{10^{2016}}-\frac{-2017}{10^{2016}}+\frac{-2017}{10^{2017}}+\frac{2016}{10^{2017}}\)
\(=\frac{1}{10^{2016}}-\frac{1}{10^{2017}}>0\)
Nên A > B
So sánh 2 phân số
A=\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)
B=\(\frac{2015+2016+2017}{2016+2017+2018}\)
Ta có : \(B=\frac{2015+2016+2017}{2016+2017+2018}\) \(=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2016}\)
Cộng vế theo vế, ta có :
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
SO SÁNH A VÀ B:
A = \(\frac{2017^{2016}}{1+2017+2017^2+...+2017^{2016}}\)
B = \(\frac{2016^{2017}}{1+2016+2016^2+...+2016^{2016}}\)
1. \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
So sánh \(B\) với \(\frac{1}{4}\)
2. SO sánh \(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\) và \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
Bài 1:
ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{100^2}\)
\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
Bài 2:
ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Học tốt nhé bn !!
a. So sánh 2 phân số sau bằng nhiều cách: 5/6 và 6/7
b. So sánh phân số sau;
A= \(\frac{2^{2015}+1}{2^{2016}+1}\)và B= \(\frac{2^{2016}+1}{2^{2017}+1}\)
So sánh 2 phân số \(A=\frac{2^{2015}+1}{2^{2016}+1}\) và \(B=\frac{2^{2016}+1}{2^{2017}+1}\)
Ta có 2A= 2(2^2015 + 1)/ 2^2016 + 1 = 2^2016 +2 / 2^2016 +1 = 2^2016+1/2^2016+1 + 1/2^2016 +1= 1 + 1/2^2016
2B= 2( 2^2016 + 1/ 2^2017+ 1) = 2^2017 +2 / 2^2017 +1 = 2^2017+1/2^2017+1 + 1/2^2017 +1 = 1 + 1/2^2017
Do 1/2^2016 > 1/2^2017 => 2A>2B => A>B
10.A=\(10.A=\frac{10.\left(2^{2015+1}\right)}{2^{2016}+1}=\frac{2^{2016+10}}{2^{2016}+1}=1+\frac{2016}{2^{2016}+1}\)
\(10.B=\frac{10.\left(2^{2016}+1\right)}{2^{2017}+1}=\frac{2^{2017}+10}{2^{2017}+1}=1+\frac{2016}{2^{2017}+1}\)
Ta có:\(\frac{2016}{2^{2016}+1}>\frac{2016}{2^{2017}+1}\)
\(A=\frac{2^{2015}+1}{2^{2016}+1}\)
\(2A=\frac{2^{2016}+1+1}{2^{2016}+1}\)
\(2A=1+\frac{1}{2^{2016}+1}\)
\(B=\frac{2^{2016}+1}{2^{2017}+1}\)
\(2B=\frac{2^{2017}+1+1}{2^{2017}+1}\)
\(2B=1+\frac{1}{2^{2017}+1}\)
=> 2A > 2B
=> A>B