\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}=\frac{0,33.x}{2009}\)

\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)

\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)

\(\Leftrightarrow\frac{1}{3}\cdot\frac{99}{100}=\frac{0,33.x}{2009}\)

\(\Leftrightarrow\frac{33}{100}=\frac{0,33.x}{2009}\)

\(\Leftrightarrow x=\frac{0,33\times100}{0,33}=100\)

A = \(\frac{\left(1.2.3......99\right)\left(1.2.3......99\right)}{\left(1.2.3......99\right)\left(2.3.4.....100\right)}=\frac{1.1}{1.100}=\frac{1}{100}\)

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{100}\right)=\frac{1}{3}\left(\frac{99}{100}\right)=\frac{33}{100}\)

33/100 nha ban that do

Vì 1=\((1)^{2}\)=\((1)^{3}\)=....

Nên tất cả các tích ở B đều có giá trị bằng 1

=> B = 1.1.1.1....1 = 1

Ta thấy mỗi tổng trên là tích của hai số tự nhiên liên tiếp. 
\(a_1=1.2\Rightarrow3a_1=1.2.3\)\(\Rightarrow3a_1=1.2.3-0.1.2\).
\(a_2=2.3\Rightarrow3a_2=2.3.3\)\(\Rightarrow3a_2=2.3.4-1.2.3\).
.....
\(a_{99}=99.100\Rightarrow3a_{99}=3.99.100\)\(\Rightarrow3a_{99}=98.99.100-97.98.99\).
Ta có:
\(3A=1.2.3+2.3.3+3.4.3+....+99.100.3\)
\(=\)\(1.2.3-0.1.2+2.3.4-1.2.3+........+98.99.100-97.98.100\)
\(=98.99.100\)
Suy ra: \(A=\frac{98.99.100}{3}=323400\).

B=1.2+2.3+3.4+...+99.100

⇒3B=1.2.3+2.3.3+....+99.100.3

⇒3B=1.2.3+2.3.(4−1)+...+99.100.(101−98)

⇒3B=1.2.3+2.3.4−1.2.3+...+99.100.101−98.99.100

⇒3B=99.100.101

\(⇒\)

\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{99.99}{98.100}\)

\(A=\left(\frac{2.3....99}{1.2....98}\right).\left(\frac{2.3....99}{3.4....100}\right)\)

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

\(A=\frac{198}{100}\)

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

\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{100-98}{98.99.100}\right)=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{18000}\)

=1-/2-1/3+1/2-1/3-1/4+1/5-1/6-1/7+1/6-1/7-1/8-.........-1/98-1/99-1/100

=1-1/100

=99/100