\(A=\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}=\frac{1.\left(100-2\right)+2\left(100-3\right)+3\left(100-4\right)+...+98\left(100-99\right)}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{1.100-1.2+2.100-2.3+3.100-3.4+...+98.100-98.99}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{\left(1.100+2.100+3.100+...+98.100\right)-\left(1.2+2.3+3.4+...+98.99\right)}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{100\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+98.99}-1\)

Ta có: 1+2+3+...+98=98.99:2=4851

Đặt B=1.2+2.3+3.4+...+98.99  => 3B=1.2.3+2.3.3+3.4.3+...+98.99.3 = 1.2.3+2.3.(4-1)+3.4(5-2)+...+98.99(100-97)

=> 3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99 = 98.99.100

=> B=33.98.100. Thay vào A được:

\(A=\frac{100.4851}{33.98.100}-1=\frac{3}{2}-1=\frac{1}{2}\)

Đặt \(A_n=1\cdot2+2\cdot3+3\cdot4+...+n\cdot\left(n+1\right)\)

Như vậy thì \(3A_n=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+n\left(n+1\right)\left[n+2-\left(n-1\right)\right]=n\left(n+1\right)\left(n+2\right)\)

Do đó \(A_n=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

Gọi số phải tính là S, ta có:

\(S=\frac{1\cdot98+2\cdot97+3\cdot96+...+98\cdot1}{1\cdot2+2\cdot3+3\cdot4+...+98\cdot99}\)

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

\(S=\frac{100\cdot\left(1+2+3+...+98\right)-A_{98}}{A_{98}}=\frac{100\cdot99\cdot49}{A_{98}}-1=\frac{100\cdot99\cdot49}{98\cdot99\cdot100:3}-1=\frac{3}{2}-1=\frac{1}{2}\)

Vậy dãy trên có giá trị là \(\frac{1}{2}\)

A =  \(\frac{1x98+2x97+3x96+...+98x1}{1x2+2x3+3x4+...+98x99}\)

A = \(\frac{1x\left(100-2\right)+2x\left(100-3\right)+3x\left(100-4\right)+...+98x\left(100-99\right)}{1x2+2x3+3x4+...+98x99}\)

A =\(\frac{1x100-1x2+2x100-2x3+3x100-3x4+...+98x100-98x99}{1x2+2x3+3x4+...+98x99}\)

A =\(\frac{100x\left(1+2+3+...+98\right)}{1x2+2x3+3x4+...+98x99}\)  - 1

Ta có: 1 + 2 + 3 + ... + 98

        = 98 x 99 : 2

        =    9702  : 2

        =           4851

Đặt B        = 1 x 2 + 2 x 3 + 3 x 4 + ... + 98 x 99

Suy ra 3B = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + ... + 98 x 99 x 100 - 97 x 98 x 99

                 = 98 x 99 x 100

B               = 98 x (99 : 3) x 100

B               = 98 x      33   x 100

Thay vào A được:

A = \(\frac{100x4851}{33x98x100}\) - 1

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

A =          \(\frac{3}{2}\)           - \(\frac{2}{2}\)

A =                            \(\frac{1}{2}\)

Vậy A bằng \(\frac{1}{2}\)

Đáp số: \(\frac{1}{2}\)

đây mà là toán lớp 4 sao(ngất)