=2016*97+2016*2+2016*1

=2016*[97+2+1]

=2016*100

=201600

2016 x 97 + 2016 x 2 + 2016

= 2016 x (97 + 2 + 1)

= 2016 x 100

= 201600

\(A< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{99.100}\)

\(A< \frac{1}{4}-\frac{1}{100}\)

\(A< \frac{6}{25}< \frac{1}{4}\)

các bạn giải mau lên

ai trả lời không mình sắp đi học rồi

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

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

= -{-4031-0-0}

=-4031

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

Do đó:

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

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

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

ngu