Question

Выполните задание.
Ответ:16/х^4 -1

Answer 1

$=( \frac{3}{y^2} - \frac{2}{x} )( \frac{3}{y^2} + \frac{2}{x}):( \frac{4}{x^2} - \frac{3^2}{y^4} )+ \frac{16}{x^4} = \\ \\ \frac{3x-2y^2}{xy^2} *\frac{3x+2y^2}{xy^2} : \frac{4y^4-9x^2}{x^2y^4} + \frac{16}{x^4} = \\ \\ \frac{9x^2-4y^4}{x^2y^4} * \frac{x^2y^4}{4y^4-9x^2} + \frac{16}{x^4} =-1+\frac{16}{x^4} =\frac{16}{x^4} -1$