Có \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}=+....+\frac{1}{99}-\frac{1}{100}\)

\(=\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

= \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right):\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=1\)

ta có:\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+...+\frac{1}{100}\)

\(\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{100}=2012\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)

bài toán được viết lại như sau:

\(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right).x=2012\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)

\(\Rightarrow x=2012\left(\frac{1}{51}+...+\frac{1}{100}\right):\left(\frac{1}{51}+...+\frac{1}{100}\right)\)

\(\Rightarrow x=2012\)

vậy x=2012

 có k không

= 1/1-1/2+1/2-1/3+1/3-...-1/100

= 1 - 1/100

= 99/100

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Sai đề à bạn    

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+....+\dfrac{1}{99\cdot100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}\right)-\dfrac{1}{100}\)

\(A=1+0-\dfrac{1}{100}\)

\(A=1-\dfrac{1}{100}< 1\)

\(\Rightarrow A< 1\)

A=1⋅21​+2⋅31​+3⋅41​+....+99⋅1001​

�=1−12+12−13+13−14+...+199−1100A=1−21​+21​−31​+31​−41​+...+991​−1001​

�=1+(−12+12−13+13−14+...+199)−1100A=1+(−21​+21​−31​+31​−41​+...+991​)−1001​

�=1+0−1100A=1+0−1001​

�=1−1100<1A=1−1001​<1

$\Rightarrow A<1$

A<1

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

=\(1-\frac{1}{100}\)

=\(\frac{99}{100}\)

A = 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100

A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/99 - 1/100

A = 1 - 1/100

A = 99/100

A=1/1x2+1/2x3+...+1/99x100

A=1-1/2+1/2-1/3+1/3-...+1/99-1/00

A=1-1/100

A=99/100

A=1 - 1/2 + 1/2 - 1/3 +...+ 1/99 - 1/100

A=1 - 1/100

A=100/100 - 1/100

A=99/100