Question

Số tự nhiên  thỏa mãn \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{16}{34}\) là 

Kèm theo cách giải

 

Answer 1

ta nhân vế trái vs 2:

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

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

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

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

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

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

\(\Rightarrow9x=-1\)

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

Answer 2

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