-1nha ko phai -11 đâu

đăt biểu thức trên là A taco

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

\(2A=2^{2016}-2^{2015}-....2^2-2\)

\(2A-A=\)\(2^{2016}-2^{2015}-....2^2-2\)-\(-\)\(2^{2015}-2^{2014}-...-2-1\)

A=\(2^{2016}-1\)

1).x:2+x:3=5/2

=>x.1/2+x.1/3=5/2

=>x(1/2+1/3)=5/2

=>x.5/6=5/2

=>x=5/2:5/6

=>x=3

1+(-2)+3+(-4)+...+2013+(-2014)+2015

=[(1-2)+(3-4)+...+(2013-2014)]+2015

=[-1+(-1)+....+(-1)]+2015

Vì từ 1 đến 2014 có 2014 số số hạng => có 1007 cặp => có 1007 số (-1)

=(-1) x 1007 +2015

=-1007+2015=1008

\(C=\dfrac{2014\left(2015^2+2016\right)-2016\left(2015^2-2014\right)}{2014\left(2013^2-2012\right)-2012\left(2013^2+2014\right)}\)

\(=\dfrac{2.2014.2016+2014.2015^2-2016.2015^2}{2014.2013^2-2012.2013^2-2.2012.2014}\)

\(=\dfrac{2.\left(2015+1\right)\left(2015-1\right)-2.2015^2}{2.2013^2-2.\left(2013+1\right)\left(2013-1\right)}\)

\(=\dfrac{2.\left(2015^2-1\right)-2.2015^2}{2.2013^2-2.\left(2013^2-1\right)}=\dfrac{-2}{2}=-1\)

\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)

=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)

\(B=2^{2016}-2^{2015}+2^{2014}-2^{2013}+.......+2^2-2^1+2^0\)

\(\Rightarrow2B=2^{2017}-2^{2016}+2^{2015}-2^{2014}+.........+2^3-2^2+2^1\)

\(\Rightarrow2B+B=3B=2^{2017}+2^0\)

\(\Rightarrow B=\frac{2^{2017}+2^0}{3}=\frac{2^{2017}+1}{3}\)