\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{100^2}-1\right)\)( có 2013 thừa số ) 

\(A=\left(-\frac{3}{2^2}\right).\left(-\frac{8}{3^2}\right).\left(-\frac{15}{4^2}\right).....\left(-\frac{\text{4056196}}{2014^2}\right)\)

\(-A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{4056196}{2014^2}=\frac{1.3.2.4.3.5....2013.2015}{2.2.3.3.4.4.....2014.2014}\)

\(-A=\frac{\left(1.2.3...2013\right).\left(3.4.5.6...2015\right)}{\left(2.3.4.5....2014\right).\left(2.3.4.5...2014\right)}=\frac{1.2015}{2.2014}=\frac{2015}{4028}\)

\(A=-\frac{2015}{4028}\)

Vậy.....

-A=(\(1-\frac{1}{2^2}\)) . (\(1-\frac{1}{3^2}\))......(\(1-\frac{1}{2014^2}\))

-A= \(\frac{3}{4}\). \(\frac{8}{9}\). ...... \(\frac{4056195}{4056196}\)

-A= \(\frac{1.3.2.4.......2013.2015}{2.2.3.3.......2.14.2014}\)

-A= \(\frac{\left(1.2.3...2013\right)\left(3.4.5...2015\right)}{\left(2.3.4...2014\right)\left(2.3.4...2014\right)}\)

-A= \(\frac{2015}{2014.2}\)

-A=\(\frac{2015}{4028}\)

mình quên A=\(\frac{-2015}{4028}\)

Ta có:\(\left(x-1\right)\left(x+1\right)=x\left(x-1\right)+x-1^2=x^2-x+x-1=x^2-1\)

Áp dụng:\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)

                  \(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot...\cdot\frac{2014^2-1}{2014\cdot2014}\)

                  \(=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot...\cdot\frac{2013\cdot2015}{2014^2}\)

                  \(=\frac{1}{2}\cdot\frac{2015}{2014}=\frac{2015}{4028}\)

có dạng \(1-\frac{1}{a^2}=\frac{\left(a-1\right)\left(a+1\right)}{a^2}\) rút gon hết còn \(\frac{1}{4028}\)

1−1a2=(a−1)(a+1)a2 rút gọn \(\frac{1}{4082}\)

Ta có

\(A=\frac{\left(1^2-2^2\right)\left(1^2-3^2\right).....\left(1^2-2014^2\right)}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)

\(\Leftrightarrow A=\frac{\left(-1\right)3\left(-2\right)4.....\left(-2013\right)2015}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)

\(\Leftrightarrow A=\frac{\left[\left(-1\right)\left(-2\right)...\left(-2013\right)\right]\left(3.4.5...2015\right)}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)

\(\Leftrightarrow A=\frac{\left(-1\right)2015}{2014.2}=-\frac{2015}{4028}< -\frac{2014}{4028}=-\frac{1}{2}\)

=> A<-1/2

gap A len 1/2

\(2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{2015}\)

\(\Rightarrow2A-A=1-\left(\frac{1}{2}\right)^{2014}\Rightarrow A=1-\left(\frac{1}{2}\right)^{2014}< 1\)

các bạn trả lời nhanh nha

\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)tương tự những cái kia rồi triệt tiêu còn phân thức đầu vs cuối

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)

\(=\frac{1}{x}-\frac{1}{x+2014}=\frac{2014}{x\left(x+2014\right)}\)

Hội con bò chúc bạn học tốt!