Theo mình bạn không nên hỏi những bài thế này vì nó rất easy

\(A=2^{2017}-\left(2^{2016}+2^{2015}+...+2^2+2^1+2^0\right)\)

\(A=2^{2017}-\left(2^0+2^1+2^2+...+2^{2015}+2^{2016}\right)\)

\(B=\left(2^0+2^1+2^2+...+2^{2015}+2^{2016}\right)\)

\(2B=\left(2^1+2^2+2^3+...+2^{2016}+2^{2017}\right)\)

\(B=\left(2^1+2^2+2^3+...+2^{2016}+2^{2017}\right)-\left(2^0+2^2+2^3+...+2^{2016}+2^{2017}\right)\)

\(B=2^{2017}-2^0=2^{2017}-1\)

\(A=2^{2017}-\left(2^{2017}-1\right)=2^{2017}-2^{2017}+1=1\)

tk mk , mk tk 

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

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

= -{-4031-0-0}

=-4031

2S=1.2+2.2²+3.2³+...+2016.2²⁰¹⁶

2S-S=S=(1.2+2.2²+...+2016.2²⁰¹⁶)-(1+2.2+...+2016.2²⁰¹⁵)

S=2016.2²⁰¹⁶-(1+2+...+2²⁰¹⁵)

S=2016.2²⁰¹⁶-(2²⁰¹⁶-1)             (1+2+...+2²⁰¹⁵=2²⁰¹⁶-1)

S=2015.2²⁰¹⁶+1

Vậy S>2015.2²⁰¹⁶

S > 2015.2^2016

trung ạ, hk hành ra rk thì đk đăng báo cụng nỏ vinh hạnh ci mô, chậc chậc, tau ns rk thôi, để m hiểu, nhìn lại mk đi, nếu mi cứ hỏi thì cụng chẳng liên quan đến tau, chỉ có mi ms tự kiểm điểm lại đk, mi lừa đk thầy, đk cô nhưng ko thể tự lừa dối mi mô

đấy, ns rk thui, suy nghị đi, đọc cho kị vô

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

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

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

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

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

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

\(\frac{B}{A}=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}+\frac{1}{2017}}\)

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

\(\frac{B}{A}=\frac{\frac{2017}{1}+\frac{2017}{2}+...+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)

\(\frac{B}{A}=\frac{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}=2017\div\frac{1}{2017}=4068289\)

đáp án=0 ( mk ko chắc)

\(a_0+a_1+...+a_{2016}=P(1)=0\)

P(x) có giá trị bằng tổng các hệ số của nó khi x=1 tức là: \(P\left(x\right)=\left(1-\frac{1}{2}-\frac{1}{2}\right)^{1008}=0\Leftrightarrow a_{2016}+a_{2015}+...+a_0=0\)