\(A=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\frac{1}{99\cdot101}\right)\)

\(A=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{10000}{99\cdot101}\)

\(A=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)\cdot...\cdot\left(100\cdot100\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(99\cdot101\right)}\)

\(A=\frac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}\)

\(A=\frac{100\cdot2}{1\cdot101}\)

\(A=\frac{200}{101}\)

Câu 1 bị sai đề bài.

Câu 2:

\(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}=\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}\)

\(=1-\frac{1}{2012}+1-\frac{1}{2013}+1+\frac{1}{2011}+\frac{1}{2011}\)

Vì:

\(\frac{1}{2011}>\frac{1}{2012};\frac{1}{2011}>\frac{1}{2013}\Rightarrow\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}>0\)

\(\Rightarrow\)\(\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}>3\)

\(\Rightarrow\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}>3\)

Tính 2 mũ76 - 2 mũ 74/ 2 mũ 78- 2 mũ 76

bài này khó lắm đó

Hình như sai đề rồi.

1/1 > 2010/2011 rồi mà!

Nếu không sai đề thì không tìm được x

Vì ta có 1 - 1/2010 = 0/2010 = 0 nên suy ra biểu thức A = 0

A=\(\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right)...\left(1-\frac{2010}{2010}\right)\left(1-\frac{2011}{2010}\right)\)

A=\(\frac{2009}{2010}.\frac{2008}{2010}...0.\frac{-1}{2010}\)

A=0