Question

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

Answer 1

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

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

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

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

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

ответ: тождество не верно