6.B=1.3.6+3.5.6+5.7.6+...+95.97.6+97.99.6

6.B=1.3.(5+1)+3.5.(7-1)+5.7.(9-3)+...+95.97.(99-93)+97.99(101-95)

6.B=1.3.5+1.3+3.5.7-1.3.5+5.7.9-3.5.7+...+95.97.99-93.95.97+97.99.101-95.97.99=1.3+97.99.101

B=(3+97.99.101)/6

  A=1.3+3.5+5.7+...+95.97+97.99

6A=1.3.6+3.5.6+5.7.6+...+95.97.96+97.99.96

    =1.3.(5+1)+3.5.(7-1)+...+95.97.(99-93)+97.99.(101-95)

    =1.1.3+1.3.5-1.3.5+3.5.7-....-95.97.99+97.99.101

    =3.97.99.101

=>A=\(\frac{3+97.99.101}{6}=\frac{1+97.33.101}{2}\)\(=161651\)

Chép sai đề bài

the minh do lai ban

D=1+3-5-7+9+11....-397-399

=3.(3/1.3+3/3.5+3/5.7+...+3/95.97+3/97.99)

=3(1-1/3+1/3-1/5+1/5-1/7+...+1/95-1/97+1/97-1/99)

=3[(1-1/99)+(1/5-1/5)+(1/7-1/7)+...+(1/97-1/97)]

=3(1-1/99)=3(99/99-1/99)=3.98/99=1.98/33=98/33

Neu la 3 ma ko phai la 3^2 thi sao : Tinh gium minh nha .

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

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

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

\(A=\frac{1}{2}.\frac{98}{99}\)

\(A=\frac{49}{99}\)

\(A=\frac{1}{1\cdot3} +\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{95\cdot97}+\frac{1}{97\cdot99}\)

\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{95\cdot97}+\frac{2}{97\cdot99}\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)

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

\(2A=\frac{98}{99}\)

\(A=\frac{98}{99}\text{ : }2\)

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

\(A=\frac{49}{99}\)

\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+.....+\frac{1}{97\cdot99}\)

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

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

\(2A=\frac{98}{99}\)

\(A=\frac{49}{99}\)

\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)

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

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

\(=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)

S=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{95.97}+\frac{1}{97.99}\)

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

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

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

S=\(\frac{49}{99}\)

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

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

= \(\frac{1}{2}\). (\(1-\frac{1}{99}\))

= \(\frac{1}{2}\). \(\frac{98}{99}\) = \(\frac{49}{99}\)

Giải:

M=\(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{95.97}+\dfrac{3}{97.99}\) 

M=\(\dfrac{3}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{95.97}+\dfrac{2}{97.99}\right)\) 

M=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\right)\) 

M=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\) 

M=\(\dfrac{3}{2}.\dfrac{32}{99}\) 

M=\(\dfrac{16}{33}\) 

Chúc bạn học tốt!

 M= 3/3.5 + 3/5.7 + 3/7.9 +.......+ 3/95.97 + 3/97.99

=3.1/2 ( 2/3.5+...+2/97.99)

=3.1/2(1/3- 1/5+...+1/97+1/99)

=3.1/2(1/3- 1/99)

=(3/2).(32/99)

=96/891

n/xét

3/3.5=(3/3-3/5).1/2

3/5.7=(3/5-3/7).1/2

...

3/97.99=(3/97-3/99).1/2

vậy M=(3/3-3/5).1/2+(3/5-3/7).1/2+...+(3/97-3/99).1/2

⇒M=1/2.(3/3-3/7+3/5-3/7+...+3/97-3/99)

      =1/2.(3/3-3/99)

     =1/2.32/33

   M =16/33

VẬY M=16/33

A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/95 - 1/97 + 1/97 - 1/99

A = 1/3 - 1/99

A = 32/99

BẠN TICK CHO MÌNH NHA

32/99

A = \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{95.97}+\dfrac{2}{97.99}\)

A = \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\)

A = \(\dfrac{1}{3}-\dfrac{1}{99}\)

A = \(\dfrac{32}{99}\)

Chúc bạn học tốt !