So sánh
\(\dfrac{2^{2015}+3^2-1}{2^{2012}+1}\) và \(\dfrac{2^{2017}+2^2}{2^{2015}+1}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
MM
1
H
16 tháng 12 2015
Có nhiều cách giải bài này. Hiện tôi có cách giải như sau tôi nghĩ là nó là ngắn nhất
Đặt: (2^2015)+1/(2^2012)+1 là A và (2^2017)+1/(2^2014)+1 là B
1/8A=(2^2015)+1/(2^2015)+8=(2^2015)+8-7/(2^2015)+8=1-7/(2^2015)+8
1/8B=(2^2017)+1/(2^2017)+8=(2^2017)+8-7/(2^2017)+8=1-7/(2^2017)+8
Vì 2^2015+8<2^2017+8 nên 7/(2^2015+8)>7/(2^2017)+8 nên 1-7/(2^2015)+8<1-7/(2^2017)+8 từ đó suy ra B>A hay 2^2017+1/(2^2014)+1>(2^2015)+1/(2^2012)+1
MT
0
MT
2
18 tháng 3 2018
\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
DT
17 tháng 4 2017
đặt phân số trên là A
tử là
(1+2015/2)+...+(1+2/2015)+(1+1/2016)+1
=2017/2+....+2017/2015+2017/2016+2017/2017
=2017.(1/2+...+1/2015+1/2016+1/2017)
=>A=\(\dfrac{2017.\left(\dfrac{1}{2}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
Vậy A=2017
LT
1
4 tháng 3 2020
Giả sử A=\(\frac{2^{2015}+1}{2^{2012}+1}\)
-->\(\frac{1}{2^3}A=\frac{2^{2015}+1}{2^{2015}+8}\)
\(\frac{1}{8}A=\frac{2^{2015}+1}{2^{2015}+1}+\frac{2^{2015}+1}{7}\)
\(\frac{1}{8}A=1+\frac{2^{2015}+1}{7}\)
B=\(\frac{2^{2017}+1}{2^{2014}+1}\)
\(\frac{1}{2^3}B=\frac{2^{2017}+1}{2^{2017}+8}\)
\(\frac{1}{8}B=\frac{2^{2017}+1}{2^{2017}+1}+\frac{2^{2017}+1}{7}\)
\(\frac{1}{8}B=1+\frac{2^{2017}+1}{7}\)
Vì \(1+\frac{2^{2015}+1}{7}< 1+\frac{2^{2017}+1}{7}\)
nên \(\frac{1}{8}A< \frac{1}{8}B\)
-->A<B
-->\(\frac{2^{2015}+1}{2^{2012+1}}< \frac{2^{2017+1}}{2^{2014}+1}\)
LT
1
10 tháng 12 2017
đặt \(A=\frac{2^{2015}+1}{2^{2012}+1}\); \(B=\frac{2^{2017}+1}{2^{2014}+1}\)
ta có :\(A=\frac{2^{2015}+1}{2^{2012}+1}\)
\(\frac{1}{2^3}A=\frac{2^{2015}+1}{2^{2015}+8}=\frac{2^{2015}+8-7}{2^{2015}+8}=1-\frac{7}{2^{2015}+8}\)
\(B=\frac{2^{2017}+1}{2^{2014}+1}\)
\(\frac{1}{2^3}B=\frac{2^{2017}+1}{2^{2017}+8}=\frac{2^{2017}+8-7}{2^{2017}+8}=1-\frac{7}{2^{2017}+8}\)
vì 22015 + 8 < 22017 + 8 nên \(\frac{7}{2^{2015}+8}>\frac{7}{2^{2015}+8}\)
\(\Rightarrow1-\frac{7}{2^{2015}+8}< 1-\frac{7}{2^{2017}+8}\)
hay \(\frac{1}{2^3}A< \frac{1}{2^3}B\)
\(\Rightarrow A< B\)
\(\frac{2^{2015}+3^2-1}{2^{2012}+1}=\frac{2^{2015}+8}{2^{2012}+1 }=\frac{2^3(2^{2012}+1) }{2^{2012}+1} =2^3=8\)
\(\frac{2^{2017}+2^2}{2^{2015}+1}=\frac{2^2(2^{2015}+1) }{2^{2015}+1} =2^2=4\)
8>4
=>....