Số tự nhiên x thỏa mãn 1/1.3+1/3.5+1/5.7+...+1/X(X+2)=16/34 là 15.

\(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{x\left(x+2\right)}=\frac{16}{34}\)
\(\frac{1}{2}.\left(\frac{1}{1.3}+....+\frac{1}{x\left(x+2\right)}\right)=\frac{16}{34}\)
\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{16}{34}\)
\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{x+2}\right)=\frac{16}{34}\)
\(\frac{1}{1}-\frac{1}{x+2}=\frac{16}{34}:\frac{1}{2}\)
\(\frac{1}{1}-\frac{1}{x+2}=\frac{16}{17}\)
\(\frac{1}{x+2}=\frac{1}{1}-\frac{16}{17}=\frac{1}{17}\Rightarrow x+2=17\Rightarrow x=15\)

là 15 

bạn Hằng Phạm làm đúng rồi 

là câu ở bài 3 violympic toán lớp 7 2016-2017

Bạn có thể chỉ cách làm không vì mình tìm được kết quả rồi

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

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{X\left(X+2\right)}\)

\(\frac{1}{2}.\left(\frac{1}{1.3}+...+\frac{1}{X\left(X+2\right)}\right)\)= \(\frac{16}{34}\)

\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{X}-\frac{1}{X+2}\right)\)

=15

TA CÓ :    1/1.3 + 1/3.5 + 1/5.7 +... + 1/X(X+2) = 8/17

        =>    2/1.3 + 2/3.5 + 2/5.7 +... + 2/X(X+2) = 8/17 . 2 = 16/17

      <=>                       1 - 1/X+2                      = 16/17

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

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

           => X+2 -1 =16 VÀ X+2 = 17

           => X = 15

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

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

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

=> \(\frac{1}{x+2}=\frac{13}{17}\)

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

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{7}\)

\(\Leftrightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{16}{7}\)

\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{7}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{16}{7}\)

\(\Rightarrow\frac{1}{x+2}=-\frac{9}{7}\)

\(\Rightarrow-9\left(x+2\right)=7\)

\(\Rightarrow x+2=-\frac{7}{9}\)

\(\Rightarrow x=-\frac{25}{9}\)

Vậy \(x=-\frac{25}{9}\)