Question

Помогите решить:
1)а²+5а-4/16-a² + 2a/8a+2a²
2)4x²/2x-3y + 12xy/3y-2x + 9y²/2x-3y

Answer 1

$\frac{a^2+5a-4}{16-a^2}+ \frac{2a}{8a+2a^2}= \frac{a^2+5a-4}{(4-a)(4+a)}+ \frac{2a}{2a(4+a)}= \frac{a^2+5a-4+4-a}{(4-a)(4+a)}= \frac{a^2+4a}{(4-a)(4+a)}= \\ \\ \frac{a(a+4)}{(4-a)(4+a)}= \frac{a}{4-a}$

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