8D= 1.3.5.8 - 3.5.7.8 + 5.7.9.8 - ... + 97.99.101.8

8D=1.3.5.(7+1)-3.5.7.(9-1)+5.7.9.(11-3) - ... + 97.99.101.(103-95)

8D=1.3.5.7+3.5-3.5.7.9-1.3.5.7+5.7.9.11-3.5.7.9-...+97.99.101.103-95.97.99.101

8D=3.5+97.99.101.103=99900024

D=12487503

\(E=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

\(E=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(E=\frac{1}{1}-\frac{1}{99}\)

\(E=\frac{98}{99}\)

E= \(\frac{2}{1.3}.\frac{2}{3.5}+...+\frac{2}{97.99}\)
E = 1 - \(\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
E = 1 - 1/99
E = 98 / 99
Chúc bạn học tốt 

\(\frac{2.6.10+6.10.14+10.14.18+...+194.198.202}{1.3.5+3.5.7+...+97.99.101}\)

\(=\frac{2^3.1.3.5+2^3.3.5.7+2^3.97.99.101}{1.3.5+3.5.7+...+97.99.101}\)

\(=\frac{2^3\left(1.3.5+3.5.7+...+97.99.101\right)}{1.3.5+3.5.7+...+97.99.101}\)

\(=\frac{2^3}{1}=8\)

Vậy A = 8

\(2S=\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{97}-\frac{2}{99}\)

\(2S=2-\frac{2}{99}\)

\(2S=\frac{196}{99}\)

\(S=\frac{196}{99}\cdot\frac{1}{2}=\frac{98}{99}\)

Ta có: S=2/1.3+2/3.5+...+2/97.99

S= 2/2.(1-1/3+1/3-1/5+...+1/97-1/99)

S= 1-1/99=98/99

\(S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

\(2S=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)

\(\Rightarrow S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(S=1-\frac{1}{99}\)

\(S=\frac{98}{99}\)

chán đời à kb đi

Có thể hướng làm là phân tích + rút gọn đấy

\(\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)-x\)\(=\frac{-100}{99}\)

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

\(\frac{98}{99}-x=\frac{-100}{99}\)

\(x=\frac{98}{99}-\left(-\frac{100}{99}\right)\)

\(x=\frac{198}{99}=2\)

CHÚC BN HOK TỐT!

ĐÚNG THÌ K CHO MK NHA!

(...) là mở đóng ngoặc đơn nha

(2/1.3 + 2/3.5 + ... + 2/97.99) - x = -100/99

=>( 1/1 - 1/3 + 1/3 - 1/5 + ... + 1/97 + 1/99) - x = -100/99

=>( 1/1 - 1/99) - x = -100/99

=>98/99 - x = -100/99

=>x = 98/99 - (-100/99)

=>x = 198/99 = 2

n=\(\frac{2}{3}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

n=\(\frac{2}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

n=\(\frac{2}{3}\left(1-\frac{1}{99}\right)\)

n=\(\frac{2}{3}\times\frac{98}{99}\)

n=\(\frac{196}{297}\)

Câu \(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{2}{99.100}\)Bạn viết \(\frac{3}{99.100}=\frac{2}{99.100}\)mik sửa lại nhé. 

\(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.100}\)

\(M=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{100-99}{99.100}\)

\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(M=\frac{3}{2}.\frac{99}{100}=\frac{297}{200}\)

\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{97.99}\)

\(N=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+....+\frac{99-97}{97.99}\)

\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)\)

\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)\)

\(\Rightarrow N=\frac{3}{2}.\frac{98}{99}=\frac{49}{33}\)

Ta thấy : \(\frac{297}{200}>\frac{49}{33}\Rightarrow M>N\)

đặt \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\) là A

A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)

2A = 2 . (\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\))

2A = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

2A = \(\frac{1}{3}-\frac{1}{99}\)

2A = \(\frac{33}{99}-\frac{1}{99}\)

2A = \(\frac{32}{99}\)

A = \(\frac{32}{99}:2\)

A = \(\frac{32}{99}.\frac{1}{2}\)

A = \(\frac{32}{198}\)

A = \(\frac{16}{99}\)

HỌC TỐT