15 баллов. сократите дробь.

Решите задачу:

$\displaystyle \frac{3x^{2}-11x-4}{12+x-x^{2}}= \frac{3(x-4)(x+ \frac{1}{3})}{(x+3)(x-4)}= \frac{3x+1}{x+3}; \\ \\3x^{2}-11x-4=0 \\ D=b^{2}-4ac=121+48=169 \\ \\ x_{1} = \frac{-b+ \sqrt{D} }{2a}=24:6=4 \\ \\ x_{2}= \frac{-b- \sqrt{D}}{2a}=-2:6=- \frac{1}{3} \\ \\ -x^{2} +x+12=0 \\ D=b^{2}-4ac=1+48=49 \\ \\ x_{1}= \frac{-b+ \sqrt{D} }{2a}=6:(-2)=-3 \\ \\ x_{2}= \frac{-b- \sqrt{D}}{2a}=-8:(-2) = 4$

$\displaystyle \frac{2-5x-2y+5xy}{10x^{3}-9x^{2}+2x}= \frac{2(1-y)-5x(1-y)}{x(10 x^{2} -9x+2)}= \frac{(2-5x)(1-y)}{10x(x-0,5)(x-0,4)}= \\ \\ =-\frac{(5x-2)(1-y)}{x(2x-1)(5x-2)}=- \frac{1-y}{x(2x-1)}; \\ \\10 x^{2} -9x+2=0 \\ D= b^{2}-4ac=81-80=1 \\ \\ x_{1}= \frac{-b+ \sqrt{D} }{2a}=10:20=0,5 \\ \\x_{2}= \frac{-b- \sqrt{D}}{2a}=8:20=0,4$