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

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