Ta có : 

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2016}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2015}{2016}\)

\(A=\frac{2.3.4.....2015}{2.3.4.....2015}.\frac{1}{2016}\)

\(A=\frac{1}{2016}\)

Vậy \(A=\frac{1}{2016}\)

Chúc bạn học tốt ~ 

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)..\left(1-\frac{1}{2016}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2015}{2016}\)

\(\Rightarrow A=\frac{1.2.3..2015}{2.3.4..2016}\)

\(\Rightarrow A=\frac{1}{2016}\)

k đúng cho mk nha

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+\right)\left(x+3\right)}+...+\frac{1}{\left(x+2015\right)\left(x+2016\right)}=\frac{1}{x+2016}\)

\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+2015}-\frac{1}{x+2016}=\frac{1}{x+2016}\)

\(\frac{1}{x}-\frac{1}{x+2016}=\frac{1}{x+2016}\)

\(\frac{1}{x}-\frac{1}{x+2016}-\frac{1}{x+2016}=0\)

\(\frac{1}{x}-\frac{2x}{x+2016}=0\)

\(\frac{x+2016}{x\left(x+2016\right)}-\frac{2x}{x\left(x+2016\right)}=0\)

\(\frac{x+2016-2x}{x\left(x+2016\right)}=0\Leftrightarrow2016-x=0\Leftrightarrow x=2016\)

\(A=\left(\frac{1}{1^2}-1\right)\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)\)

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

suy ra A>B

\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{2}{3}\right)...\left(1-\frac{2015}{2016}\right)\)

\(=\frac{1}{2}\cdot\frac{1}{3}...\frac{1}{2016}\)

\(=\frac{1}{1\cdot2...2016}\)

\(\Rightarrow2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2014}\)

\(\Rightarrow2A-A=A=1-\left(\frac{1}{2}\right)^{2015}\)

Với B tương tự nhưng là lấy 3B