Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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
-->\(\frac{2^{2015}+1}{2^{2012+1}}< \frac{2^{2017+1}}{2^{2014}+1}\)
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 \(\frac{1}{2^3}A< \frac{1}{2^3}B\)
\(\Rightarrow A< B\)
\(\frac{2^{2017}+1}{2^{2014}+1}>1\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{2^{2017}+\left(1+3\right)}{2^{2014}+\left(1+3\right)}\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{2^{2017}+4}{2^{2014}+4}\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{4\left(2^{2015}+1\right)}{4\left(2^{2012}+1\right)}\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{2^{2015}+1}{2^{2012}+1}\)
