Question Number 116507 by bemath last updated on 04/Oct/20
Answered by bobhans last updated on 04/Oct/20
![let (1/((1/x)+(1/2))) = r ⇒ (1/((2+x)/(2x)))= r ⇒((2x)/(2+x)) = r . the equation equivalent to (1/([(1/(r+r))+(1/(r+r))])) = (x/(36)) ⇒(1/(((1/(2r))+(1/(2r))))) = (x/(36)) ⇒ (1/(((1/r)))) = (x/(36)) ; r = (x/(36)) ⇒((2x)/(2+x)) = (x/(36)) ; x≠0 ⇒ (2/(2+x)) = (1/(36)) ⇒x=70](Q116508.png)
$$\mathrm{let}\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{2}}}\:=\:\mathrm{r}\:\Rightarrow\:\frac{\mathrm{1}}{\frac{\mathrm{2}+\mathrm{x}}{\mathrm{2x}}}=\:\mathrm{r} \\ $$$$\Rightarrow\frac{\mathrm{2x}}{\mathrm{2}+\mathrm{x}}\:=\:\mathrm{r}\:.\:\mathrm{the}\:\mathrm{equation}\:\mathrm{equivalent} \\ $$$$\mathrm{to}\:\frac{\mathrm{1}}{\left[\frac{\mathrm{1}}{\mathrm{r}+\mathrm{r}}+\frac{\mathrm{1}}{\mathrm{r}+\mathrm{r}}\right]}\:=\:\frac{\mathrm{x}}{\mathrm{36}}\:\Rightarrow\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{2r}}+\frac{\mathrm{1}}{\mathrm{2r}}\right)}\:=\:\frac{\mathrm{x}}{\mathrm{36}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{r}}\right)}\:=\:\frac{\mathrm{x}}{\mathrm{36}}\:;\:\mathrm{r}\:=\:\frac{\mathrm{x}}{\mathrm{36}} \\ $$$$\Rightarrow\frac{\mathrm{2x}}{\mathrm{2}+\mathrm{x}}\:=\:\frac{\mathrm{x}}{\mathrm{36}}\:;\:\mathrm{x}\neq\mathrm{0} \\ $$$$\Rightarrow\:\frac{\mathrm{2}}{\mathrm{2}+\mathrm{x}}\:=\:\frac{\mathrm{1}}{\mathrm{36}}\:\Rightarrow\mathrm{x}=\mathrm{70} \\ $$