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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
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{7}\)
\(\Leftrightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{16}{7}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{7}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{16}{7}\)
\(\Rightarrow\frac{1}{x+2}=-\frac{9}{7}\)
\(\Rightarrow-9\left(x+2\right)=7\)
\(\Rightarrow x+2=-\frac{7}{9}\)
\(\Rightarrow x=-\frac{25}{9}\)
Vậy \(x=-\frac{25}{9}\)
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
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
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
Ta có : \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{16}{34}\)
=> \(2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}\right)=2.\frac{16}{34}\)
=> \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{16}{17}\)
=> \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)
=> \(1-\frac{1}{x+2}=\frac{16}{17}\)
=> \(\frac{1}{x+2}=1-\frac{16}{17}=\frac{1}{17}\)
=> \(x+2=17\)
=> \(x=15\)
=>1/1-1/3+1/3-1/5+1/5-1/7+....+1/x-1/(x+2)=16/34
=>1/1-1/(x+2)=16/34
=>1/(x+2)=1-16/34
=>1/(x+2)=9/17
=>(x+2).9=17
=>(x+2)=17/9
=>x=17/9-2
=>x=-1/9(không là số tự nhiên)
vậy không có số tự nhiên x thoả mãn điều kiện bài toán
