a)Đặt \(L=\frac{1}{2^{2015}}+\frac{1}{2^{2014}}+...+\frac{1}{2^0}\)
\(2L=\left(1+\frac{1}{2}+...+\frac{1}{2^{2015}}\right)\)
\(2L=2+1+...+\frac{1}{2^{2014}}\)
\(2L-L=\left(2+1+...+\frac{1}{2^{2014}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2015}}\right)\)
\(2L=2-\frac{1}{2^{2015}}\) thay vào ta có:
\(B=\frac{1}{2^{2016}}-\left(2-\frac{1}{2^{2015}}\right)=\frac{1}{2^{2016}}-2+\frac{1}{2^{2015}}\)
b)Ta có:\(\begin{cases}\left|x+1\right|\ge0\\\left|x+4\right|\ge0\end{cases}\)\(\Rightarrow\left|x+1\right|+\left|x+4\right|\ge0\)
\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow3x\ge0\Rightarrow x\ge0\)
Với \(x\ge0\) ta có\(x+1+x+4=3x\)
\(\Rightarrow2x+5=3x\Rightarrow x=5\) (thỏa mãn)
Vậy x=5
