Solve the following equation:

$$
\frac{3+2 x}{1+2 x}-\frac{5+2 x}{7+2 x}=1-\frac{4 x^{2}-2}{7+16 x+4 x^{2}}
$$
$$
\frac{3+2 x}{1+2 x}-\frac{5+2 x}{7+2 x}=1+\frac{2}{1+2 x}-\frac{5+2 x}{7+2 x}=1+\frac{9-8 x-4 x^{2}}{7+16 x+4 x^{2}}
$$

thus, our equation can also be written as:

$$
1-\frac{4 x^{2}+8 x-9}{7+16 x+4 x^{2}}=1-\frac{4 x^{2}-2}{7+16 x+4 x^{2}}
$$

which means

$$
8 x-9=-2
$$

from which

$$
x=\frac{7}{8}
$$

(Jenö Silbermann, Nagyvárad.)