Question

Помогите пожалуйста с алгеброй

Answer 1

$1)\frac{42-x-x^{2}}{2x^{2}-13x+6 }=\frac{-(x^{2}+x-42) }{2x^{2}-13x+6 } =\frac{-(x+7)(x-6)}{2(x-6)(x-\frac{1}{2})}=\frac{-(x+7)}{2(x-\frac{1}{2})}=\frac{-(x+7)}{2x-1}=\frac{x+7}{1-2x}$

$2)\frac{9x^{3}-9x^{2}+2x}{1-3x+y-3xy}=\frac{x(9x^{2}-9x+2)}{(1+y)-(3x+3xy)}=\frac{9x(x-\frac{1}{3})(x-\frac{2}{3})}{(1+y)-3x(1+y)}=\frac{x(3x-1)(3x-2)}{(1+y)(1-3x)}=-\frac{x(3x-2)}{1+y}=\frac{x(2-3x)}{1+y}$

$3)\frac{3x^{2}-11x-4 }{12-x-x^{2}}=\frac{3(x-4)(x+\frac{1}{3})}{-(x^{2}+x-12)}=\frac{3(x-4)(x+\frac{1}{3})}{-(x+4)(x-3)}=\frac{(x-4)(3x+1)}{(x+4)(3-x)}$

$4)\frac{9a-4}{a+7}-\frac{44-16a}{a^{2}+5a-14 }=\frac{9a-4}{a+7}-\frac{44-16a}{(a+7)(a-2)}=\frac{9a^{2}-18a-4a+8-44+16a}{(a+7)(a-2)}=\frac{9a^{2}-6a-36 }{(a+7)(a-2)}=\frac{3(3a^{2}-2a-12)}{(a+7)(a-2)}$