Đăt A=1.3+5.7+9.11+...+97.99
      B=3.5+7.9+11.13+...+99.101
Ta có: A+B=1.3+3.5+5.7+7.9+...+97.99+99.101
6(A+B)=1.3.6+3.5.6+5.7.6+…+97.99.6+99.101.6
=1.3.(5+1)+3.5.(7−1)+5.7(9−3)+…+97.99(101−95)+99.101.(103−97)
=1.3.5+1.3+3.5.7−1.3.5+5.7.9−3.5.7+…+97.99.101−95.97.99+99.101.103−97.99.101
=3+99.101.103=1029900
⇒A+B=171650
Lại có: 
B−A=3.(5−1)+7.(9−5)+11.(13−9)+...+99.(101−97)
=4.(3+7+11+...+99)
=4.(3+99).252=5100
Suy ra ta có: {A+B=171650B−A=5100⇔ {A=83275B=88375
Vây 1.3+5.7+9.11+...+97.99=83275

b, B= 1.2.3+2.3.4+3.4.5+...+98.99.100

4B = (1.2.3+2.3.4+3.4.5+...+98.99.100).4

4B=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+.....+98.99.100.(101-97)

4B= 1.2.3.4-0.1.2.3 + 2.3.4.5-1.2.3.4+......+98.99.100.101-97.98.99.100

4B=98.99.100.101

B=98.99.25.101

B=6999300

a) \(A=1.99+2.98+3.97+...+50.50\)

\(A=1.\left(100-1\right)+2.\left(100-2\right)+3.\left(100-3\right)+...+50.\left(100-50\right)\)

\(A=100.\left(1+2+3+...+50\right)-\left(1^2+2^2+3^2+...+50^2\right)\)

Xét \(1+2^2+3^2+...+50^2\)

\(=1.\left(2-1\right)+2.\left(3-1\right)+3\left(4-1\right)+...+50.\left(51-1\right)\)

\(=\left(1.2+2.3+3.4+...+50.51\right)-\left(1+2+3+...+50\right)\)

\(=\frac{1.2.3+2.3.\left(4-1\right)+...+50.51.\left(52-49\right)}{3}-1275\)

\(=\frac{1.2.3-1.2.3+2.3.4-...-49.50.51+50.51.52}{3}-1275\)

\(=44200-1275\)

\(=42925\)

Thay vào A ta được:

\(A=127500-42925=84575\)

Ta thấy: 
1/1 + 1/99 = (99+1)/(1.99)=100/(1.99) 
1/3 + 1/97 = (97+3)/(3.97)=100/(3.97) 
1/5 + 1/95 = (95+5)/(5.95)=100/(3.97) 
… 
1/97 + 1/3 = (3+97)/(97.3)=100/(97.3) 
1/99 + 1/1 = (1+99)/(99.1)=100/(99.1) 
=> 
1/(1.99)=(1/1+1/99)/100 
1/(3.97)=(1/3+1/97)/100 
… 
1/(99.1)=(1/99+1/1)/100 
------------------------------ cộng 2 vế của các đẳng thức trên. Ta được đẳng thức: 

1/(1.99) + 1/(3.97)+ 1/(5.95) +...+ 1/(97.3) + 1/(99.1 ) 
=[(1/1+1/99)+(1/3+1/99)+…+(1/99+1/1)]/1... 
=2(1+1/3+1/5+1/7…+1/99]/100 
=(1+1/3+1/5+1/7…+1/99]/50 
Vậy: 
A=(1+1/3+1/5+1/7+...+1/97+1/99) / [ 1/(1.99) + 1/(3.97)+ 1/(5.95) +...+ 1/(97.3) + 1/(99.1 ) ] 
A=(1+1/3+1/5+1/7+...+1/97+1/99)/[(1+1/3... 
A=50. 

ko chắc nhé

bn ko lam theo cach lop 5 dc a . sao ma dan the

A=618

Cho A=(1+1/3+1/5+...+1/97+1/99)/(1/1*99+1/3*97+1/5*95+...+1/49*51) rut gon ta duoc A=3

click cho tui nhen chac chan do

=1/2

bn cho mk xem lai cach lam nha

=100 mới đúng

 \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}}\)

=  \(\frac{100\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}{100\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}\right)}\)

=  \(\frac{100\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}{\frac{1+99}{1.99}+\frac{3+97}{3.97}+\frac{5+95}{5.95}+...+\frac{49+51}{49.51}}\)

= \(\frac{100\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}{\left(\frac{1}{1}+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)}\)

=  \(\frac{100\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{51}+...+\frac{1}{99}}\)

= 100