$\frac{x^2}{x-2}:\frac{x^2+2x}{4-x^2}=$

$=\frac{x^2}{x-2}·\frac{4-x^2}{x^2+2x}=$

$=\frac{x^2}{x-2}·\frac{\left(2+x\right)\left(2-x\right)}{x\left(x+2\right)}$

$\frac{x^2}{x-2}:\frac{x^2+2x}{4-x^2}=$

$=\frac{x\bcancel{{}^2}}{\bcancel{x-2}}·\frac{-\bcancel{\left(x+2\right)}\bcancel{\left(x-2\right)}}{\bcancel{x}\bcancel{\left(x+2\right)}}=$

$=-x$

$x\in \mathrm{ℝ}\setminus \left\{-2;0;2\right\}$

$\frac{3x+13}{3x^2+19x+26}:\frac{x+7}{x^2+2x}=$

$=\frac{3x+13}{3x^2+19x+26}·\frac{x^2+2x}{x+7}=$

$=\frac{3x+13}{\left(3x+13\right)\left(x+2\right)}·\frac{x\left(x+2\right)}{x+7}$

$\frac{3x+13}{3x^2+19x+26}:\frac{x+7}{x^2+2x}=$

$=\frac{\bcancel{3x+13}}{\bcancel{\left(3x+13\right)}\bcancel{\left(x+2\right)}}·\frac{x\bcancel{\left(x+2\right)}}{x+7}=$

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

$x\in \mathrm{ℝ}\setminus \left\{-7;-\frac{13}{3};-2;0\right\}$

$\frac{1-9x^2}{x^4+2x^2+1}:\frac{18x^3+12x^2+2x}{x^4-1}=$

$=\frac{1-9x^2}{x^4+2x^2+1}·\frac{x^4-1}{18x^3+12x^2+2x}=$

$=\frac{\left(1-3x\right)\left(1+3x\right)}{{\left(x^2+1\right)}^2}·\frac{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}{2x{\left(3x+1\right)}^2}$

$\frac{1-9x^2}{x^4+2x^2+1}:\frac{18x^3+12x^2+2x}{x^4-1}=$

$=\frac{\left(1-3x\right)\bcancel{\left(3x+1\right)}}{\left(x^2+1\right)\bcancel{{}^2}}·\frac{\bcancel{\left(x^2+1\right)}\left(x+1\right)\left(x-1\right)}{2x{\left(3x+1\right)}^{\bcancel{2}}}=$

$=\frac{\left(1-3x\right)\left(x+1\right)\left(x-1\right)}{2x\left(x^2+1\right)\left(3x+1\right)}$

$x\in \mathrm{ℝ}\setminus \left\{-1;-\frac{1}{3};0;1\right\}$