Question

Найдите значение дроби 1) x²-8x-33/10x+30 при x = -9;12;111

Answer 1

$\frac{x^2-8x-33}{10x+30} = \frac{x(x-8)-33}{10(x+3)}$
при x=-9 : $\frac{-9(-9-8)-33}{10(-9+3)} = \frac{120}{60} =2$
при x=12 $\frac{12(12-8)-33}{10(12+3)} = \frac{15}{150} =0.1$
при x=111 :$\frac{111(111-8)-33}{10(111+3)} = \frac{11400}{1140} =10$