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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
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)
\(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
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
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
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 :
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
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 !!
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:\(\frac{2016}{2^{2016}+1}>\frac{2016}{2^{2017}+1}\)
Mình sẽ dạy bạn cách rút gọn để làm các bài khác với cách tương tự
Ta thấy các số 2016 và 2017 ở p/s A cứ lặp đi lặp lại 3 lần . Mà các c/s 2 , 0 , 1 , 6 cứ 3 c/s lặp lại 1 lần , các c/s ko hề đc gấp 2, gấp 3,...
=> 201620162016 = 2016 . 100010001 ( 3 số 0 là khoảng cách các c/s , 3 số 1 là 3 số 2016
=> 201720172017 = 2017 . 100010001
=> A \(\frac{201620162016}{201720172017}=\frac{2016.100010001}{2017.100010001}=\frac{2016}{2017}=B\)
=> A = B ( Mình lớp 6 nhé )
\(\frac{201620162016}{201720172017}=\frac{2016}{2017}\)