\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+....+\frac{9}{98.99}+\frac{9}{99.100}\)

\(A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=9\left(1-\frac{1}{100}\right)\)

\(A=9\cdot\frac{99}{100}=\frac{891}{100}\)

\(A=9\left(\frac{1}{1x2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

=> \(A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

=> \(A=9\left(1-\frac{1}{100}\right)=\frac{9.99}{100}=\frac{891}{100}\)

=> A=8,91

Ta có:\(A=\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{98.99}+\frac{9}{99.100}\)

\(=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(=9\left(1-\frac{1}{100}\right)\)

\(=9.\frac{99}{100}=\frac{891}{100}\)

A = 1.2+2.3+3.4+......+99.100 
Gấp A lên 3 lần ta có: 
A . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3 
A . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98) 
A . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100 
A . 3 = 99.100.101 
A = 99.100.101 : 3 
A = 33.100.101 
A = 333 300

bạn giải được không

3A=3.(1x2+2x3+3x4+....+99x100)

3A=3x1x2x+3x2x3+3x3x4+...+3x99x100

3A=(3-0)x1x2+(4-1)x2x3+(5-2)x3x4+...+(101-98)x99x100

3A=1x2x3-0+2x3x4-1x2x3+3x4x5-2x3x4+...+99x100x101-98x99x100

3A=99x100x101-0

3A=99x100x101

A=33x100x101

A=333 300

A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100

A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3 A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98) ..................................

A x 3 = 99x100x101 A = 333300

A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
..................................
A x 3 = 99x100x101
A = 333300

Đặt A=1.2+2.3+3.4+...+99.100

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)

3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

3A=99.100.101

A=333300

A = 333 300 nhé bn

333300

**** mik nha bak

A = 1.2 + 2.3 + ... + 99.100

          3A = 1.2.3 + 2.3.(4-1) +...+ 99.100.(101-98)

          3A = 1.2.3 + 3.2.4 - 1.2.3 + ... + 99.100.101 - 98.99.100

          3A = 99.100.101

          3A = 999900

A = 3A : 3 = 999900 : 3 = 333300

A = 1.2 + 2.3 + ... + 99.100

3A = 1.2.3 + 2.3.(4-1) + ... + 99.100.(101-98)

3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 99.100.101 - 98.99.100

3A = 99.100.101

3A = 999900

A = 333300

anh chi lam gioi qua

B = \(\dfrac{2}{1\times2}\) + \(\dfrac{2}{2\times3}\)+ \(\dfrac{2}{3\times4}\)+...+ \(\dfrac{2}{99\times100}\)

B = 2 \(\times\) ( \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\)+ \(\dfrac{1}{3\times4}\)+....+ \(\dfrac{1}{99\times100}\))

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

B = 2 \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\))

B = 2 \(\times\) \(\dfrac{99}{100}\)

B = \(\dfrac{99}{50}\)