\(S=1-2+2^2-2^3+...+2^{2012}-2^{2013}\)

\(\Rightarrow2S=2-2^2+2^3-2^4+...+2^{2013}-2^{2014}\)

\(\Rightarrow2S+S=2-2^2+2^3-...-2^{2014}+1-2^2-2^3+...-2^{2013}\)

\(\Rightarrow3S=1-2^{2014}\)\(\Rightarrow3S-2^{2014}=1-2^{2015}\)

\(2^{x+1}\cdot2^{2014}=2^{2015}\\ 2^{x+1}=2^{2015}:2^{2014}\\ 2^{x+1}=2\\ =>x+1=1\\ x=1-1\\ x=0\)

\(B=2^{2018}-2^{2017}-2^{2016}-2^{2015}-2^{2014}\)

\(=>2B=2^{2019}-2^{2018}-2^{2017}-2^{2016}-2^{2015}\)

\(=>2B+B=2^{2019}-2^{2014}\)

\(=>B=\dfrac{2^{2019}-2^{2014}}{3}\)

ta có: \(S=1-2+2^2-2^3+2^4-2^5+...+2^{2013}-2^{2014}\)

\(\Rightarrow2S=2-2^2+2^3-2^4+2^5-2^6+...+2^{2014}-2^{2015}\)

=> 2S + S = -2²⁰¹⁵ + 1

=> 3S = -2²⁰¹⁵ + 1

=> 3S - 1 = -2²⁰¹⁵

=> 1 - 3S = 2²⁰¹⁵

( cn về S = 1 - 2 + 22 - 23 + 24-25+...+22013 - 22014 mk vx chưa hiểu quy luật của nó lắm, thật lòng xl bn nha! mk chỉ bk z thoy!)

a/ Điều kiện xác định \(\hept{\begin{cases}a^2+a\ne0\\a^2-a\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}a\ne0\\a\ne1\\a\ne-1\end{cases}}}\)

b/ \(M=\frac{a^2-1}{2016+2015a^2}\left(\frac{2015a-2016}{a+a^2}+\frac{2016+2015a}{a^2-a}\right)\)

\(=\frac{\left(a-1\right)\left(a+1\right)}{2016+2015a^2}\left(\frac{2015a-2016}{a\left(a+1\right)}+\frac{2016+2015a}{a\left(a-1\right)}\right)\)

\(=\frac{\left(a-1\right)\left(a+1\right)}{2016+2015a^2}\left(\frac{2015a-2016}{a\left(a+1\right)}+\frac{2016+2015a}{a\left(a-1\right)}\right)\)

\(=\frac{\left(a-1\right)\left(a+1\right)}{2016+2015a^2}.\frac{2\left(2015a^2+2016\right)}{a\left(a+1\right)\left(a-1\right)}\)

\(=\frac{2}{a}=\frac{2}{2016}=\frac{1}{1008}\)

Bài 1:

Ta có: \(3n+1⋮n-1\)

\(\Leftrightarrow3n-3+4⋮n-1\)

mà \(3n-3⋮n-1\)

nên \(4⋮n-1\)

\(\Leftrightarrow n-1\inƯ\left(4\right)\)

\(\Leftrightarrow n-1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{2;0;3;-1;5;-3\right\}\)(tm)

Vậy: \(n\in\left\{2;0;3;-1;5;-3\right\}\)

\(a^2+b^2+c^2=1\Rightarrow\left\{{}\begin{matrix}a^2\le1\\b^2\le1\\c^2\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|a\right|\le1\\\left|b\right|\le1\\\left|c\right|\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3\le a^2\\b^3\le b^2\\c^3\le c^2\end{matrix}\right.\)

\(\Rightarrow a^3+b^3+c^3\le a^2+b^2+c^2=1\)

Đẳng thức xảy ra khi và chỉ khi: \(\left(a;b;c\right)=\left(0;0;1\right)\) và các hoán vị

\(\Rightarrow S=0+0+1=1\)

Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{2015a}{2015c}=\frac{2016b}{2016d}\)

                                       \(=\frac{2015a-2016b}{2015c-2016d}=\frac{2015a+2016b}{2015c+2016d}\)

\(\Rightarrow\frac{2015a-2016b}{2015a+2016b}=\frac{2015c-2016d}{2015c+2016d}\)(đpcm)

Từ \(\frac{a}{b}=\frac{c}{d}\)ta suy ra:

\(\frac{a}{b}=\frac{c}{d}=\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}=\frac{a-b}{a+b}=\frac{c-d}{c+d}\Rightarrow\frac{2015a-2016b}{2015a+2016b}\)\(=\frac{2015c-2016d}{2015c+2016d}\)(Áp dụng tính chất dãy tỉ số bằng nhau)