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\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{x^2}\right)=\frac{1007}{2012}\)
\(\Rightarrow\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{\left(x^2-1\right)}{x^2}=\frac{1007}{2012}\)
\(\Rightarrow\frac{1.3.2.4.3.5...\left(x-1\right)\left(x+1\right)}{\left(2.3.4..x\right)^2}=\frac{1007}{2012}\)
\(\Rightarrow\frac{\left[2.3.4...\left(x-1\right)\right].\left[3.4.5...\left(x+1\right)\right]}{\left(2.3.4....x\right)\left(2.3.4....x\right)}=\frac{1007}{2012}\)
\(\Rightarrow\frac{\left(x+1\right)}{2x}=\frac{1007}{2012}\)
\(\Rightarrow2002.\left(x+1\right)=1007.2x\)
\(\Rightarrow2012x+2012=2014x\)
\(\Rightarrow2x=2012\)
\(\Rightarrow x=1006\)
Vậy x = 1006
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
gọi biểu thức 2^4+2^8+....+2^2016 là A ta có
A=2^4+2^8+.....+2^2016
8A=2^4+2^8+.....+2^2010
8A-A=2-2^2010
7A=1+2-2^2010
![](https://rs.olm.vn/images/avt/0.png?1311)
2A=2+1+1/2+...+1/22011
=>2A-A=2+1+1/2+...+1/22011-(1+1/2+1/22+...+1/22012)
=>A=2-1/22012
Vậy ...