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.

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

\(=2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)

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

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

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

2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2009.2011

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2009 - 1/2011

= 1 - 1/2011

= 2010/2011

Ta có : \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

Đặt : \(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)

\(A-\frac{2}{1\cdot3}=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)

\(2A-\frac{2}{1\cdot3}=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-...+\frac{2}{99}-\frac{2}{101}\)

\(2A-\frac{2}{3}=\frac{2}{3}-\frac{2}{101}\)

\(2A-\frac{2}{3}=\frac{196}{303}\)

\(A-\frac{2}{3}=\frac{98}{303}\)

\(A=\frac{98}{303}+\frac{2}{3}=\frac{100}{101}\)

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)

\(=\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+...+\frac{101-99}{99\times101}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{1}-\frac{1}{101}\)

\(=\frac{100}{101}\)

A = 2/1x3 + 2/3x5 + 2/5x7 + ... + 2/99x101

A = 2/1 - 2/101 = 200/101

Kết quả là 200/101 bạn nhé

2/2 + 1x3 / 3x5 + 2/2 + ······ + 5x7 / 97x99 + 2 / 99x101 
= 1-1 / 3 + ​​1 / 3-1 / 5 + 1 / 5-1 / 7 + ... ... + 1 / 97-1 / 99 + 1 / 99-1 / 101 
= 1-1 / 101 
= 100/101

chưa chắc là 200/101

\(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+....+\frac{2}{59x61}\)

\(=\frac{2}{2}x\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+.....+\frac{2}{59x61}\right)\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{59}-\frac{1}{61}\)

\(=1-\frac{1}{61}=\frac{60}{61}\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\)

\(=1-\frac{1}{2015}=\frac{2014}{2015}\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\)

\(=1-\frac{1}{2015}\)

\(=\frac{2015}{2015}-\frac{1}{2015}\)

\(=\frac{2014}{2015}\)