Dựa vào 2/n(n+1)(n+2)= 1/n(n+1) - 1/(n+1)(n+2)

Ta có

Z = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/98.99.100

2Z =  2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ... + 2/98.99.100

2Z = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/98.99 - 1/99.100

2Z = 1/1.2 - 1/99.100

2Z = 4949/9900

=> Z = 4949/19800

=> 4949/19800 . x = 49/200

                           x = 49/200 : 4949/19800

                           x = 99/101

Vậy x = 99/101

Ủng hộ nha

 Minh moi hoc tieu hoc thoi

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tich nha

1/1.2.3 + 1/2.3.4 +....+1/98.99.100

= 1/2 . (3-1/1.2.3 + 4-2/2.3.4 +....+ 100-98/98.99.100)

= 1/2 . (3/1.2.3 -1/1.2.3 + 4/2.3.4 - 2/2.3.4 +.......+ 100/98.99.100 - 98/98.99.100)

= 1/2 . (1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 +......+ 1/98.99 - 1/99.100)

= 1/2 . (1/2 - 1/9900)

= 1/2 . 4949/9900

= 4949/19800

A=11.2.3+12.3.4+13.4.5+...+198.99.100=11.2−12.3+12.3−13.4+...+198.99−199.100=11.2−199.100=494919800

=1/1.2.3+1/2.3.4+1/3.4.5+............+1/98.99.100

 $=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)$

$=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)$

$=\frac{1}{2}\cdot \frac{4949}{9900}$

$=\frac{4949}{19800}$

cho mình xin lỗi vì đáp án mình gửi lên nó bị lỗi nhá

B=1/1.2.3+1/2.3.4+1/3.4.5+............+1/98.99.100

 \(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(=\frac{1}{2}\cdot\frac{4949}{9900}\)

\(=\frac{4949}{19800}\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

\(B=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)

\(B=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(B=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}\)

Ta xét:

\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3};\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4};...;\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)

Qua công thức trên, bạn có thể rút ra tổng quát: (đây là mình nói thêm)

\(\frac{1}{n.\left(n+1\right)}-\frac{1}{\left(n+1\right).\left(n-2\right)}=\frac{2}{n.\left(n+1\right).\left(n+2\right)}\)

Ta suy ra:

\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)

       \(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

      Thấy \(-\frac{1}{2.3}+\frac{1}{2.3}=0;-\frac{1}{3.4}+\frac{1}{3.4}=0;...\)

\(\Rightarrow2B=\frac{1}{2}-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{9900}=\frac{4950}{9900}-\frac{1}{9900}=\frac{4949}{9900}\)

\(\Rightarrow B=\frac{4949}{9900}:2=\frac{4949}{19800}\)

Mình nhầm, công thức tổng quát mình nói thêm bạn đổi cái n-2 thành n+2 nha

 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ............. + 1/ 98.99.100