\(2^{2016}+4^{2016}+6^{2016}+...+20^{2016}=2^{2016}\left(1+2^{2016}+3^{2016}+...+10^{2016}\right)\)

Do đó:

\(A=\frac{1^{2016}+2^{2016}+3^{2016}+...+10^{2016}}{2^{2016}+4^{2016}+6^{2016}+...+20^{2016}}=\frac{1}{2^{2016}}\)

a, - { -(2016 +2015) - [ - (2016 - 2015) - (2016+2015) ] }

= -{-(2016+2015)-[-0-0]}

= -{-4031-0-0}

=-4031

\(A=\dfrac{10^{2016}-1+3}{10^{2016}-1}=1+\dfrac{3}{10^{2016}-1}\)

\(B=\dfrac{10^{2016}-3+3}{10^{2016}-3}=1+\dfrac{3}{10^{2016}-3}\)

mà \(10^{2016}-1>10^{2016}-3\)

nên A<B

Ta có:A=\(\frac{2016^{2016}+2}{2016^{2016}-1}\)>1

=>A<\(\frac{2016^{2016}+2-2}{2016^{2016}-1-2}\)=\(\frac{2016^{2016}}{2016^{2016}-3}\)=B

=>A<B(công thức nếu \(\frac{a}{b}\)>1 thì \(\frac{a}{b}\)<\(\frac{a-n}{b-n}\)(nEN)

CM công thức:

Ta có \(\frac{a}{b}\)>1=>a>b=>a=b+n(nEN)

Ta so sánh \(\frac{a}{b}\) và \(\frac{a-n}{b-n}\)(nEN)

Mà a*(b-n)=ab-an=ab-(b+n)*n=ab-(bn+n²)=ab-bn-n²

       b*(a-n)=ba-bn

Vì ab-bn-n²<ba-bn

=>\(\frac{a}{b}\)<\(\frac{a-n}{b-n}\)

\(A=\sqrt[]{1+2015^2+\dfrac{2015^2}{2016^2}}+\dfrac{2015}{2016}\)

\(\Leftrightarrow A=\sqrt[]{\left(1+2015\right)^2-2.2015+\dfrac{2015^2}{\left(2015+1\right)^2}}+\dfrac{2015}{2016}\)

\(\Leftrightarrow A=\sqrt[]{\left(1+2015-\dfrac{2015}{2015+1}\right)^2}+\dfrac{2015}{2016}\)

\(\Leftrightarrow A=\left|1+2015-\dfrac{2015}{2016}\right|+\dfrac{2015}{2016}\)

\(\Leftrightarrow A=1+2015-\dfrac{2015}{2016}+\dfrac{2015}{2016}\)

\(\Leftrightarrow A=1+2015=2016\)