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

2^2015+1/2^2012+1 < 2^2017+1/2^2014+1 

2²⁰¹⁵+1/2²⁰¹²+1<2²⁰¹⁷+1/2²⁰¹⁴+1...........dung 100%

Ai h mk mk se h lai

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 \(\frac{1}{2^3}A< \frac{1}{2^3}B\)

\(\Rightarrow A< B\)

\(\frac{2^{2017}+1}{2^{2014}+1}>1\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{2^{2017}+\left(1+3\right)}{2^{2014}+\left(1+3\right)}\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{2^{2017}+4}{2^{2014}+4}\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{4\left(2^{2015}+1\right)}{4\left(2^{2012}+1\right)}\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{2^{2015}+1}{2^{2012}+1}\)

Đặt

\(A=\frac{2^{2015}+1}{2^{2012}+1}\) và \(B=\frac{2^{2017}+1}{2^{2014}+1}.\)

Ta có:

\(\frac{1}{8A}=2^{2015}+\frac{1}{2^{2015}}+8=2^{2015}+8-\frac{7}{2^{2015}}+8=1-\frac{7}{2^{2015}}+8.\)

\(\frac{1}{8B}=2^{2017}+\frac{1}{2^{2017}}+8=2^{2017}+8-\frac{7}{2^{2017}}+8=1-\frac{7}{2^{2017}}+8.\)

Vì \(2^{2015}< 2^{2017}.\)

\(\Rightarrow\frac{7}{2^{2015}}>\frac{7}{2^{2017}}.\)

\(\Rightarrow\frac{7}{2^{2015}}+8>\frac{7}{2^{2017}}+8.\)

\(\Rightarrow1-\frac{7}{2^{2015}}+8< 1-\frac{7}{2^{2017}}+8.\)

\(\Rightarrow A< B.\)

Hay \(\frac{2^{2015}+1}{2^{2012}+1}< \frac{2^{2017}+1}{2^{2014}+1}.\)

Chúc bạn học tốt!

Đặt \(A=\frac{2^{2015}+1}{2^{2012}+1}\) và \(B=\frac{2^{2017}+1}{2^{2014}+1}\)

Ta có: \(\frac{1}{8A}=2^{2015}+\frac{1}{2^{2015}}+8=2^{2015}+8-\frac{7}{2^{2015}}+8=1-\frac{7}{2^{2015}}+8\)

\(\frac{1}{8B}=2^{2017}+\frac{1}{2^{2017}}+8=2^{2017}+8-\frac{7}{2^{2017}}+8=1-\frac{7}{2^{2017}}+8\)

Ta có: \(7^{2015}< 7^{2017}\)

\(\Rightarrow\frac{7}{2^{2015}}>\frac{7}{2^{2017}}\)

\(\Rightarrow1-\frac{7}{2^{2015}}+8< 1-\frac{7}{2^{2017}}+8\)

hay A<B

hay \(\frac{2^{2015}+1}{2^{2012}+1}\)<\(\frac{2^{2017}+1}{2^{2014}+1}\)

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

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