uk, cái bạn tên Phong Thần công nhận giỏi thật nha

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

\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)

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

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 \(\frac{1}{2^3}A< \frac{1}{2^3}B\)

\(\Rightarrow A< B\)