Помагите прям в край нужно плезз

№2.
$\frac{b^2}{2b-a} *(2- \frac{a}{b} ) = \frac{b^2}{2b-a} * \frac{2b -a}{b} = \frac{(b*b) *(2b-a)}{b(2b-a)}= \frac{b*1*1}{1*1}= \frac{b}{1} =b \\ \\ \frac{d^3}{c-3d} * ( \frac{c}{d} -3) = \frac{d^3}{c-3d} * \frac{c-3d}{d} = \frac{d^2}{1} =d^2 \\ \\ \frac{5+3x}{x^5 } : ( \frac{3}{5} + \frac{1}{x} ) = \frac{5+3x}{x^5 } : \frac{3x + 1*5}{5x} = \frac{5+3x}{x^5 } : \frac{5+3x}{5x} = \frac{5+3x}{x^5} * \frac{5x}{5+3x} = \frac{5}{x^4} \\ \\$
$\frac{3+2y}{y^4} : ( \frac{2}{3} + \frac{1}{y} )= \frac{3+2y}{y^4} : \frac{2y +1*3}{3y} = \frac{3+2y}{y^4} * \frac{3y}{3+2y} = \frac{3}{y^3}$