Question

(3/2x-y -2/2x+y - 1/ 2x-5y): 4y^2/x^2-y^2 решите

Answer 1

$(\frac{3}{2x-y} -\frac{2}{2x+y} - \frac{1}{2x-5y}):\frac{4y^2}{x^2-y^2}=\\\\=\frac{3(2x+y)(2x-5y)-2(2x-y)(2x-5y)-(2x-y)(2x+y)}{(2x-y)(2x+y)(2x-5y)}*\frac{(x-y)(x+y)}{4y^2} =\\\\=\frac{3(4x^2-8xy-5y^2)-2(4x^2-12xy+5y^2)-(4x^2-y^2)}{(2x-y)(2x+y)(2x-5y)}*\frac{(x-y)(x+y)}{4y^2}=\\\\=\frac{12x^2-24xy-15y^2-8x^2+24xy-10y^2-4x^2+y^2}{(2x-y)(2x+y)(2x-5y)}*\frac{(x-y)(x+y)}{4y^2}=\\\\=\frac{-24y^2(x-y)(x+y)}{(2x-y)(2x+y)(2x-5y)*4y^2}=-\frac{6(x-y)(x+y)}{(2x-y)(2x+y)(2x-5y)}$