Question

Решите плз
a)$\frac{3(x-2y)}{4(2y-x)}$
$\frac{3b(5a-6)}{6-5a}$
$\frac{x^{2}-16y^{2}} {4y+x}$

Answer 1

$\frac{3(x-2y)}{4(2y-x)}=\frac{-3(2y-x)}{4(2y-x)}=-\frac{3}{4}$

$\frac{3b(5a-6)}{6-5a}=\frac{-3b(6-5a)}{6-5a}=-3b$

$\frac{x^{2}-16y^{2}}{4y+x}=\frac{(x-4y)(x+4y)}{x+4y}=x-4y$

Answer 2

$\frac{3(x-2y)}{4(2y-6)} =\frac{3(x-2y)}{4*(-(x-2y))} =\frac{3*(-1)}{4} =\frac{-3}{4} =-\frac{3}{4} \\\\\frac{3b(5a-6)}{6-5a} =\frac{3b(5a-6)}{-(5a-6)} =3b*(-1)=-3b\\\\\frac{x^2-16y^2}{4y+x} =\frac{(x-4y)(x+4y)}{4y+x} =x-4y$