ta nhân vế trái vs 2:

\(\frac{2}{1.3}+\frac{2}{3.5}+......+\frac{2}{x\left(x+2\right)}=\frac{8}{17}\)

\(\frac{1}{ }-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{8}{17}\)

\(1-\frac{1}{x+2}=\frac{8}{17}\)

\(\Rightarrow17\left(x+1\right)=8\left(x+2\right)\)

\(\Rightarrow17x+17=8x+16\)

\(\Rightarrow17x-8x=-17+16\)

\(\Rightarrow9x=-1\)

\(\Rightarrow x=\frac{-1}{9}\)

2(1/1.3+1/3.5+1/5.7+...+1/x(x+2) )=16/34 *2

2/1.3+2/3.5+2/5.7+...+2/x(x+2)=32/34=16/17

1/1-1/3+1/3-1/5+1/5-1/7+...+1/x-1/x+2=16/17

1/1-1/x+2=16/17

1/x+2=1/1-16/17

1/x+2=1/17

suy ra x+2=17

         x=17=2=15

=> \(2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{16}{34}\)

=>\(2.\left(1-\frac{1}{x+2}\right)=\frac{16}{34}\)

=>\(1-\frac{1}{x+2}=\frac{4}{17}\)

=> \(\frac{1}{x+2}=\frac{13}{17}\)

=>\(x=-\frac{9}{13}\)

bài 1

[(x+2)/10¹⁰]+ [(x+2)/11¹¹]= [(x+2)/12¹²]+[(x+2)/13¹³]

=>[(x+2)/10¹⁰]+[(x+2)/11¹¹] - [(x+2)/12¹²]-[(x+2)/13¹³] = 0

=>(x+2).[(1/10¹⁰)+(1/11¹¹)-(1/12¹²)-(1/13¹³)=0

Vì [(1/10¹⁰)+(1/11¹¹)-(1/12¹²)-(1/13¹³)] khác 0

=>x+2=0

=>x=-2

Bài 1: x=-2

Bài 2:x=17

Bài 3:x=2014

y=2010

Bài 1 : -2

Bài 2 : 15

Bải 3 : x =2014 ; y = 2010

1/2[2/1.3+2/3.5+2/5.7+.........+2/x(x+2)]=16/34

2/1.3+2/3.5+2/5.7+......+2/x(x+2)=16/34:1/2=16/17

1/1-1/3+1/3-1/5+1/5-1/7+.....+1/x-1/x+2=16/17

1-1/x+2=16/17

1/x+2=1-16/17=1/17

suy ra:x+2=17

x=17-2

x=15

Ta có\(\frac{1}{1\cdot3}\) +\(\frac{1}{3\cdot5}\)+\(\frac{1}{5\cdot7}\)+.....+\(\frac{1}{x\cdot\left(x+2\right)}\)=\(\frac{16}{34}\)

=> 2(\(\frac{1}{1\cdot3}\)+\(\frac{1}{3\cdot5}\)+\(\frac{1}{5\cdot7}\)+......+\(\frac{1}{x+\left(x+2\right)}\)) = \(\frac{16}{34}\)*2

=>  \(\frac{2}{1\cdot3}\)+\(\frac{2}{3\cdot5}\)+\(\frac{2}{5\cdot7}\)+.....+\(\frac{2}{x\cdot\left(x+2\right)}\)= \(\frac{32}{34}\)

1-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+.....+\(\frac{1}{x}\)-\(\frac{1}{x+2}\)=\(\frac{32}{34}\)

1-\(\frac{1}{x+2}\)=\(\frac{32}{34}\)

\(\frac{1}{x+2}\)= 1-\(\frac{32}{34}\)

\(\frac{1}{x+2}\)= \(\frac{1}{17}\)

=> x+2=17

x=17-2 

x=15

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)

\(\Rightarrow\frac{1}{2}\left[\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}\right]=\frac{16}{17}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{16}{17}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{17}\Rightarrow x+2=17\Rightarrow x=15\)

\(B=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)

Ta có:

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x\left(x+2\right)}\right)=\frac{8}{17}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{16}{17}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{17}\)

\(\Rightarrow x+2=17\)

\(\Rightarrow x=15\)

Vậy \(x=15\)

xin lưu ý, h24 rất hoan nghêng các bn có cách giải khác so với ng đã giải để làm phong phú kiến thức nhân loại nhưng k chào đón cách giải giống hệt bn đã giải vì tạo sự nhàm chán

A\(A=\frac{1}{1.3}+..+\frac{1}{x\left(x+1\right)}\)

\(2A=\frac{1}{1}-\frac{1}{\left(x+1\right)}\)

\(A=\frac{x}{2.\left(x+1\right)}=\frac{8}{17}=\frac{16}{2.17}\)

X=16

17 - 1= 16

= > x = 16

 tk mình nha

17 - 1= 16

= > x = 16

 tk mình nha