Докажите тождество: 18xy/2y+3x + 1/2y-3x : ( 4/4y^2-9x^2 - 6y-9x/8y^3+27x^3)= 3x+ 3y

$\frac{18xy}{2y+3x}+\frac{\frac{1}{2y-3x}}{\frac{4}{4y^2-9x^2}-\frac{6y-9x}{8y^3+27x^3}}$

$1)\frac{4}{4y^2-9x^2}-\frac{6y-9x}{8y^3+27x^3}=\frac{4}{(2y-3x)(3x+2y)}-\frac{6y-9x}{(3x+2y)(9x^2-6xy+4y^2)}= \\ \\ = \frac{4(9x^2-6xy+4y^2)-(6y-9x)(2y-3x)}{(2y-3x)(3x+2y)(9x^2-6xy+4y^2)}= \frac{(36x^2-24xy+16y^2)-(12y^2-36xy+27x^2)}{(2y-3x)(3x+2y)(9x^2-6xy+4y^2)} = \\ \\ =\frac{9x^2+12xy+4y^2}{(2y-3x)(3x+2y)(9x^2-6xy+4y^2)} =\frac{(3x+2y)^2}{(2y-3x)(3x+2y)(9x^2-6xy+4y^2)} = \\ \\ =\frac{3x+2y}{(2y-3x)(9x^2-6xy+4y^2)}$

$2) \frac{1}{2y-3x}:\frac{3x+2y}{(2y-3x)(9x^2-6xy+4y^2)}=\frac{1}{2y-3x}*\frac{(2y-3x)(9x^2-6xy+4y^2)}{3x+2y}= \\ \\ =\frac{9x^2-6xy+4y^2}{3x+2y}$

$3) \frac{18xy}{2y+3x}+\frac{9x^2-6xy+4y^2}{3x+2y}= \frac{18xy+9x^2-6xy+4y^2}{3x+2y} = \\ \\ =\frac{9x^2+12xy+4y^2}{3x+2y} =\frac{(3x+2y)^2}{3x+2y} =3x+2y$

$3x+2y \neq 3x+ 3y$

ответ: тождество не верно