Question

помогите пожалуйста решить, 9 класс​

Answer 1

Решение:

$1)~ \dfrac{8}{6+x} - \dfrac{8}{x-7} =\dfrac{8(x-7)-8(6+x)}{(6+x)(x-7)} = \dfrac{8x-56 - 48 - 8x}{(6+x)(x-7)} =-\dfrac{104}{(6+x)(x-7)}$

$2)~\dfrac{9a^2 + y^2}{3a-y} +\dfrac{6ay}{y - 3a} = ~\dfrac{9a^2 + y^2}{3a-y} -\dfrac{6ay}{3a-y} = \\ \\=\dfrac{9a^2 + y^2-6ay}{3a-y} = \dfrac{(3a-y)^2}{3a-y} =3a-y$

$3)~ \dfrac{ay}{a-yb} + \dfrac{3a-by}{by-a} =\dfrac{ay}{a-by} - \dfrac{3a-by}{a-by} =\\ \\=\dfrac{ay-3a + by}{a-by} =\dfrac{a(y-3) + by}{a-by}$

$4)~ \dfrac{a^2x^2+36y^2}{ax-6y} + \dfrac{12axy}{6y-ax} =\dfrac{a^2x^2+36y^2}{ax-6y} - \dfrac{12axy}{ax-6y} =\\ \\= \dfrac{a^2x^2+36y^2-12axy}{ax-6y}=\dfrac{(ax-6y)^2}{ax-6y} =ax-6y$