Tìm x sao cho : 1/(1.3) + 1/(3.5) + ….... + 1/[x.(x+2)] = 16/34
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Số tự nhiên x thỏa mãn 1/1.3+1/3.5+1/5.7+...+1/X(X+2)=16/34 là 15.
\(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{x\left(x+2\right)}=\frac{16}{34}\)
\(\frac{1}{2}.\left(\frac{1}{1.3}+....+\frac{1}{x\left(x+2\right)}\right)=\frac{16}{34}\)
\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{16}{34}\)
\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{x+2}\right)=\frac{16}{34}\)
\(\frac{1}{1}-\frac{1}{x+2}=\frac{16}{34}:\frac{1}{2}\)
\(\frac{1}{1}-\frac{1}{x+2}=\frac{16}{17}\)
\(\frac{1}{x+2}=\frac{1}{1}-\frac{16}{17}=\frac{1}{17}\Rightarrow x+2=17\Rightarrow x=15\)
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