E=2.(2+2)+4.(2+4)+6.(6+2)+....+98.(98+2).

E=2.2+2.2+2.4+4.4+...+98.98+2.98

E=2.(2+4+6+...+98)+(2.2+4.4+6.6+...+98.98)

E=2.2450+40425.4

E=4900+161700

E=166600

Dấu chấm bằng dấu nhân.

Bạn T-i-c-k đúng cho mình đi mình giải thích tại sao :(2.2+4.4+…+98.98)=40425.4

S=(2+98)*(4+6)+...+100+100+102

100*10+....+100+100*102
=224400

\(B=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{46\cdot48}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{46\cdot48}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{48}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{23}{48}=\dfrac{23}{96}< \dfrac{1}{4}\)

\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{40.42}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{40}-\dfrac{1}{42}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{42}\right)\)

\(=\dfrac{1}{2}.\dfrac{10}{21}\)

\(=\dfrac{5}{21}\)

\(#Wendy.Dang\)

\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{40\cdot42}\)

\(=\dfrac{1}{2}\cdot\left(2\cdot\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{40\cdot42}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{40\cdot42}\right)\)

\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{40}-\dfrac{1}{42}\right)\)

\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{42}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{41}{42}\)

\(=\dfrac{41}{84}\)

\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}\)

\(=\frac{49}{100}\)

Ta có:

\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{98.100}\)

\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{98.100}\)

\(\Rightarrow2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\)

\(\Rightarrow2A=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(\Rightarrow A=\frac{49}{100}\div2=\frac{49}{200}\)

Vậy giá trị của biểu thức là \(\frac{49}{200}\)

\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)

= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)

= \(\frac{1}{2}-\frac{1}{100}\)

= \(\frac{49}{100}\)

Ps: Tham khảo: [Toán nâng cao 5] - Tính nhanh phân số - tth - YouTube