\(A=\frac{2^{2014}+1}{2^{2014}}=\frac{2^{2014}}{2^{2014}}+\frac{1}{2^{2014}}=1+\frac{1}{2^{2014}}\)
\(B=\frac{2^{2014}+2}{2^{2014}+1}=\frac{2^{2014}+1+1}{2^{2014}+1}=\frac{2^{2014}+1}{2^{2014}+1}+\frac{1}{2^{2014}+1}=1+\frac{1}{2^{2014}+1}\)
Ta có: \(\frac{1}{2^{2014}}>\frac{1}{2^{2014}+1}\)
\(\Rightarrow1+\frac{1}{2^{2014}}>1+\frac{1}{2^{2014}+1}\)
\(\Rightarrow\frac{2^{2014}+1}{2^{2014}}>\frac{2^{2014}+2}{2^{2014}+1}\)
\(\Rightarrow A>B\)
Tham khảo nhé ~
A= 2^2014+1/2^2014
B= 2^2014+2/2^2014+1
vì 1/2^2014<2/2^2014+1
=> A<B
cái này nhìn là bt mà ko cần chứng minh phức tạp lắm đâu bn nhìn một tí là làm dc ngay
\(A=\frac{2^{2014}+1}{2^{2014}}=\frac{2^{2014}}{2^{2014}}+\frac{1}{2^{2014}}=1+\frac{1}{2^{2014}}\)
\(B=\frac{2^{2014}+2}{2^{2014}+1}=\frac{2^{2014}+1+1}{2^{2014}+1}=\frac{2^{2014}+1}{2^{2014}+1}+\frac{1}{2^{2014}+1}=1+\frac{1}{2^{2014}+1}\)
Vì \(2^{2014}< 2^{2014}+1\Rightarrow\frac{1}{2^{2014}}>\frac{1}{2^{2014}+1}\Rightarrow1+\frac{1}{2^{2014}}>1+\frac{1}{2^{2014}+1}\) hay A > B
Vậy A > B