Question

a ) Tính B = \(\frac{1}{2^{2016}}-\left(\frac{1}{2^{2015}}+\frac{1}{2^{2014}}+...+\frac{1}{2^1}+\frac{1}{2^0}\right)\)

b ) Tìm x biết : | x + 1 | + | x + 4 | = 3x

Answer 1

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:

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

Answer 2

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

 

Answer 3

â hình như sai