Lời giải:

$A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{99-97}{97.99}$

$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}$

$=1-\frac{1}{99}=\frac{98}{99}$

a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)

Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)

\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)

\(\dfrac{1}{x+1}=\dfrac{1}{100}\)

\(\Rightarrow x+1=100\)

\(x=100-1\)

\(x=99\)

câu b thiếu kết quả đúng không bn?

Công thức: \(\dfrac{1}{a\times b}=\) 1/ khoảng cách giữa a và b \(\times\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\)

* Bạn làm theo công thức và vẫn dụng câu b nhé.

Đặt biểu thức = A

6A = 1x3x6 + 3x5x6+....+ 97x99x6

= 1x3x(1+5) + 3x5x(7-1) + 5x7x(9-3) +.....+ 97x99x(101-95)

= 1x3+1x3x5+3x5x7-1x3x5+5x7x9-3x5x7+.....+97x99x101-95x97x99

= 1x3+97x99x101

= 969906

=> A = 161651

k mk nha

2/3x5 + 2/5x7 + 2/7x9 + ......+ 2/97x99 = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99 
= 1/3 - 1/99 = 96/3.99 = 32/99 

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)

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

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

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

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

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

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

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

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

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

=  1 - 1/99

= 98/99

sao lại là 1- 1/3 + 1/3 -1/5 + ...... 1/97 - 1/99 hả bạn :|

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{95.97}\)\(+\frac{2}{97.99}\)

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

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

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

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x.\left(x+2\right)}=\frac{2015}{2016}\)

\(\Rightarrow\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{x}-\frac{2}{\left(x+2\right)}=\frac{2015}{2016}\)

\(\Rightarrow2-\frac{2}{x+2}=\frac{2015}{2016}\)

\(\Rightarrow\frac{2}{x+2}=2-\frac{2015}{2016}\)

\(\Rightarrow\frac{2}{x+2}=\frac{2017}{2016}\)

\(\Rightarrow2017.\left(x+2\right)=2.2016\)

\(\Rightarrow2017x+4034=4032\)

\(\Rightarrow2017x=-2\)

\(\Rightarrow x=-\frac{2}{2017}\)

Vậy......

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{x\cdot\left(x+2\right)}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)

\(=1-\frac{1}{x+2}=\frac{2015}{2016}\)

=>\(\frac{1}{x+2}=\frac{1}{2016}\)

=>\(x+2=2016\)

=>\(x=2014\)

Vậy.......

Ta có:

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{2015}{2016}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{2015}{2016}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{2015}{2016}=\frac{1}{2016}\)

\(\Rightarrow x+2=1\div\frac{1}{2016}=2016\)

\(\Rightarrow x=2016-2=2014\)

Vậy giá trị của x là 2014.