Trả lời:

\(S=\) \(\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\)\(\left(-3\right)^{2015}\)

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

\(-3S-S=\)\([\left(-3\right)^1+\left(-3\right)^2+...+\)\(\left(-3\right)^{2016}\)\(]\)\(-\)\([\)\(\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\)\(\left(-3\right)^{2015}\)\(]\)

\(\left(-3-1\right)S=\)\(\left(-3\right)^1+\left(-3\right)^2+...+\)\(\left(-3\right)^{2016}\)\(-\)\(\left(-3\right)^0-\left(-3\right)^1-\left(-3\right)^2-...-\)\(\left(-3\right)^{2015}\)

\(-4S=\)\(\left[\left(-3\right)^1-\left(-3\right)^1\right]\)\(+\)\(\left[\left(-3\right)^2-\left(-3\right)^2\right]\)\(+\)\(...\)\(+\)\(\left[\left(-3\right)^{2015}-\left(-3\right)^{2015}\right]\)\(+\)\(\left[\left(-3\right)^{2016}-\left(-3\right)^0\right]\)

\(-4S=\)\(0+0+...+0+\left(-3\right)^{2016}-1\)

\(-4S=\)\(3^{2016}-1\)

\(S=\frac{-3^{2016}+1}{4}\)

Vậy \(S=\frac{-3^{2016}+1}{4}\)

P/s: Không chắc có đúng ko. 

Hok tốt!

Vuong Dong Yet

Ta có : 

\(S=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{2015}\)

\(3S=\left(-3\right)^1+\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2015}\)

\(3S-S=\left[\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{2016}\right]+\left[\left(-3\right)^0+\left(-3\right)^1+...+\left(-3\right)^{2015}\right]\)

\(2S=\left(-3\right)^{2016}-\left(-3\right)^0\)

\(2S=3^{2016}-1\)

\(S=\frac{3^{2016}-1}{2}\)

Vậy \(S=\frac{3^{2016}-1}{2}\)

Chúc bạn học tốt ~ 

= (-3)²⁰¹⁶ -1

Suy ra S = 3^0 - 3^1 + 3^2+.....- 3^2015

=> 3S = 3^1 - 3^2 + 3^3 + .... - 3^2016

=> S+3S = 3^0 - 3^2016 (Phần này mình làm tắt, các bạn cố gắng hiểu nhé )

=> 4S = 1 - 3^2016

=> S = (1 - 3^2016) / 4

Phần sau các bạn tự tính nhé!

Nếu đúng thì nhé!

s = \(\left(-3\right)^{2016}-\left(-3\right)^0\)

Ta có B= (-3)⁰+ (-3)¹+.....+(-3)²⁰¹⁵

=> -3B= -3.[(-3)⁰+(-3)¹+...+(-3)²⁰¹⁵]

=> -3B= (-3)¹+ (-3)²+....+(-3)²⁰¹⁶

=> -3B-B= (-3)¹ +(-3)²+....+ (-3)²⁰¹⁶ - [(-3)⁰+(-3)¹+....+ (-3) ²⁰¹⁵

=> -4B= (-3)²⁰¹⁶- (-3)¹

=>-4B= (-3)²⁰¹⁶+ 1

=> B= (-3)²⁰¹⁶+ 1 / -4

Mình nhầm, -4B= (-3)²⁰¹⁶- (-3)⁰

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

=>-4B=(-3)^2016-1

=>\(B=\dfrac{-3^{2016}+1}{4}\)

a) \(S=1+\left(-2\right)+3+\left(-4\right)+...+\left(-2014\right)+2015\)

\(\Leftrightarrow S=\left(1-2\right)+\left(3-4\right)+....+\left(2013-2014\right)+2015\)

Vì từ 1 đến 2014 có 2014 số hạng => có 1007 cặp => Có 1007 cặp -1 và số 2015

\(\Rightarrow S=\left(-1\right)\cdot1007+2015\)

<=>S=-1007+2015

<=> S=1008

\(\left(x-1\right)\left(-x+2\right)=0\Leftrightarrow x=1;x=2\)

\(\left(x+2\right)\left(x+1-x+3\right)=0\Leftrightarrow x=-2\)

\(\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\left(x-2\right)\left(-x-2\right)=0\Leftrightarrow x=-2;x=2\)

\(i,\left(x-1\right)\left(x+3\right)-\left(x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3-2x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(-x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\ k,\left(x+2\right)\left(x+1\right)-\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x+1-x+3\right)=0\\ \Leftrightarrow4\left(x+2\right)=0\\ \Leftrightarrow x+2=0\\ \Leftrightarrow x=-2\\ l,\left(x-2\right)\left(x+3\right)=\left(x-2\right)\left(2x+5\right)\\ \Leftrightarrow\left(x-2\right)\left(2x+5\right)-\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2x+5-x-3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)