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

\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{100-99}{99.100}\)

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

\(A=1-\frac{1}{100}=\frac{99}{100}\)

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

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

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

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

Theo đề bài ta có:

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

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

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

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

Vậy \(\Rightarrow A=\frac{99}{100}\)

ta có :\(\frac{1}{1\cdot2}=\frac{1}{1}-\frac{1}{2}\) 

          \(\frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)

           \(\frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4}\)

            ......

          \(\frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)

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

\(=>A=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

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

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

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

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

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

\(\Rightarrow A< 1\)

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

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

1/1.2+1/2.3+....+1/99.100

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

=1/1-1/100=100/100-1/100=99/100

99/100

tick nha tui tick lại

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

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

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

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

Cho hai số biết rằng bớt số thứ nhất 28 đơn vị thì được số thứ hai va 1/3 số thứ nhất bằng 3/5 số thứ hai.Tìm hai số đó

sai roi

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

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

A=1/1 - 1/100

A=99/100

ta có: A=1/1x2+1/2x3+...+1/99x100

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

A=1-1/100

A=99/100

 A=1/1x2+1/2x3+.....+1/99x100

BL:

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

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

A=1/1 - 1/100

A=99/100