=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 .

2/1.3 + 2/3.5 + 2/5.7 +...+ 2/97.99 
=(1/1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/97-1/99) 
=1-1/99=98/99 

 

\(S=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)

\(=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)

\(=\dfrac{1}{1}-\dfrac{1}{11}=\dfrac{11}{11}-\dfrac{1}{11}=\dfrac{10}{11}\)

Bài này lớp 6 học rùi! 

S = 312/25

Bạn có cần giải cặn kẽ ko

x = \(\frac{2}{99}\)

\(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)-x=-\frac{100}{99}\)

\(\Rightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)-x=-\frac{100}{99}\)

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

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

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

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

Vậy x = 2

tách\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+......+\(\frac{2}{97.99}\)ra thành A

A=1/1-1/3+1/3-1/5+1/5-........+1/97-1/99

A=1-1/99=98/99

=>x=98/99-\(\frac{-100}{99}\)=2

k mình nha

tham the 

có giỏi thì làm một câu xem nào

đơn giản à