2.
Za x podstawiamy -1
[tex]A. \ L = 6(-1-4) = 6\cdot(-5) = -30 \neq P\\\\\boxed{B. \ L = (-1)^{3}-4\cdot(-1) = -1 + 4 = 3 = P}\\\\C. \ \frac{-1}{3} -\frac{-1}{4}=-\frac{4}{12}+\frac{3}{12} = -\frac{1}{12}\neq P[/tex]
Odp. B. (D. ucięte).
3.
[tex]3(2x^{2}+15)=-9-6x(1-x)\\\\6x^{2}+45=-9-6x+6x^{2}\\\\6x = -9-45\\\\6x = -54 \ \ /:6\\\\\boxed{x = -9}\\\\Odp. \ B.[/tex]
