Question

Кто может решить? ❤️❤️​

Answer 1

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

Answer 2

тут....................

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