=（2-1）*（2+1）+（4-1）*（4+1）+ ...+(2n-1)*(2n+1) =(2^2-1)+(4^2-1)+...+(4n^2-1) =(2^2+4^2+...+4n^2)-(1+1+...+1) =4(1^2+2^2+...n^2)-n n(n+1)(2n+1)/6： 1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6n^2=n 1x3+3x5+5x7+7x9+...+17x19 =4(1^2+2^2+...n^2)-n =4*n(n+1)(2n+1)/6-n； n=10，1x3+3x5+5x7+7x9+...+17x19=1530

\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{17\cdot19}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)

\(A=\frac{1}{3}-\frac{1}{19}\)

\(A=\frac{16}{57}\)

Dấu "." là dấu nhân nhá ^^

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{17\cdot19}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{17}-\frac{1}{19}\)

\(=\frac{1}{3}-\frac{1}{19}\)

\(=\frac{16}{57}\)

\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{17\times19}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)

\(=\frac{1}{3}-\frac{1}{19}\)

\(=\frac{16}{57}\)

Code : Breacker

`1/[1xx3]+1/[3xx5]+1/[5xx7]+...+1/[17xx19]`

`=1/2xx(2/[1xx3]+2/[3xx5]+....+2/[17xx19])`

`=1/2xx(1-1/3+1/3-1/5+....+1/17-1/19)`

`=1/2xx(1-1/19)`

`=1/2xx18/19`

`=9/19`

\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{17\times19}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)

=\(\frac{1}{3}-\frac{1}{19}=\frac{19}{57}-\frac{3}{57}=\frac{19-3}{57}=\frac{16}{57}\)

Ta có;\(\frac{4}{1\times3}+\frac{4}{3\times5}+\frac{4}{5\times7}+....+\frac{4}{19\times21}\)

\(=2\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+....+\frac{2}{19\times21}\right)\)

\(=2\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(=2\times\left(1-\frac{1}{21}\right)=2\times\frac{20}{21}=\frac{40}{21}\)

4/1 x 3 + 4/ 3 x 5 + 4/ 5 x 7 + ....+ 4/ 17 x 19 + 4/ 19 x 21

= 2 x ( 2/ 1 x 3 + 2/ 3 x 5 + 2/ 5 x 7 + ...+ 2/ 17 x 19 + 2/ 19 x 21 ) 

= 2 x ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ...+ 1/17 - 1/19 + 1/19 - 1/21 ) 

= 2 x ( 1 - 1/21 ) 

= 2 x  20/21

= 40/21 

Chúc bạn học giỏi !!! 

Đặt \(K=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{17.19}+\frac{4}{19.21}\)

\(\Leftrightarrow K=\left(\frac{4}{1}-\frac{4}{3}\right)+\left(\frac{4}{3}-\frac{4}{5}\right)+\left(\frac{4}{5}-\frac{4}{7}\right)+\left(\frac{4}{7}-\frac{4}{9}\right)+...+\left(\frac{4}{19}-\frac{4}{21}\right)\)

\(\Leftrightarrow K=\frac{4}{1}-\frac{4}{3}+\frac{4}{3}-\frac{4}{5}+\frac{4}{5}-\frac{4}{7}+\frac{4}{7}-\frac{4}{9}+...+\frac{4}{19}-\frac{4}{21}\)

\(\Leftrightarrow K=\frac{4}{1}-\frac{4}{21}=\frac{84}{21}-\frac{4}{21}=\frac{80}{21}\)

\(\left(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{17\times19}\right)\times57-2\times\left(x-1\right)=10\)

\(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\right)\times57-2\times\left(x-1\right)=10\)

\(\left(\frac{1}{3}-\frac{1}{19}\right)\times57-2\times\left(x-1\right)=10\)

\(\frac{16}{57}\times57-2\times\left(x-1\right)=10\)

\(16-2\times\left(x-1\right)=10\)

\(2\times\left(x-1\right)=16-10\)

\(2\times\left(x-1\right)=6\)

\(x-1=6:2\)

\(x-1=3\)

\(x=3+1\)

\(x=4\)

Nhưng bn ơiX là x^2 hay tách biệt nếu tách biệt thì là 9/49 còn nếu là x^2 thì là 3/7 nhé

\(=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)xX=\frac{9}{7} \)\(=\left(\frac{1}{3}-\frac{1}{21}\right)xX=\frac{9}{7}\)\(=\frac{2}{7}xX=\frac{9}{7}\)

\(X=\frac{9}{7}:\frac{2}{7}\)

\(X=\frac{9}{2}\)

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.