\(P=\frac{3^{2016}-6^{2016}+9^{2016}-12^{2016}+15^{2016}-18^{2016}}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)

\(=\frac{\left(1.3\right)^{2016}-\left(2.3\right)^{2016}+\left(3.3\right)^{2016}-\left(4.3\right)^{2016}+\left(5.3\right)^{2016}-\left(6.3\right)^{2016}}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)

\(=\frac{1^{2016}.3^{2016}-2^{2016}.3^{2016}+3^{2016}.3^{2016}-4^{2016}.3^{2016}+5^{2016}.3^{2016}-6^{2016}.3^{2016}}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)

\(=\frac{-3^{2016}\left(-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}\right)}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)

\(=-3^{2016}\)

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

Do đó:

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

Giả sử \(\dfrac{3^{2016}+6^{2016}}{7^{2016}+14^{2016}}=\dfrac{9^{2016}}{21^{2016}}\)

=> \(3^{2016}\cdot21^{2016}+6^{2016}\cdot21^{2016}=7^{2016}\cdot9^{2016}+14^{2016}\cdot9^{2016}\)

=\(63^{2016}+126^{2016}=63^{2016}+126^{2016}\) (giả sử đúng)

Vậy \(\dfrac{3^{2016}+6^{2016}}{7^{2016}+14^{2016}}=\dfrac{9^{2016}}{21^{2016}}\)

\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+...+\frac{2016}{501}}{\frac{-1}{1.2}+\frac{-1}{3.4}+...+\frac{-1}{999.1000}}=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{999.1000}\right)}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)

\(=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{500}\right)\right]}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+....+\frac{1}{999}+\frac{1}{1000}\right)}=\frac{2016}{-1}=-2016\)

Vậy B = - 2016

Bạn Xyz cho mik hỏi ở phần mẫu số tại sao lại có -2*(1/2+1/4+...+1/1000) vậy? Nó ở đâu ra thế?

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

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

= -{-4031-0-0}

=-4031

a)\(=\frac{2017}{2016}.\frac{3}{4}-\frac{1}{2016}.\frac{3}{4}\)

\(=\frac{3}{4}\left(\frac{2017}{2016}-\frac{1}{2016}\right)\)

\(=\frac{3}{4}.1\)

\(=\frac{3}{4}\)

b)\(=\frac{2015}{2016}\left(\frac{1}{2}+\frac{1}{3}-\frac{5}{6}\right)\)

\(=\frac{2015}{2016}.0\)

\(=0\)