Odpowiedzi

2009-10-11T19:39:47+02:00
Na cynniki
30 ab²(c-3)+ a²b(3-c) =
=30 ab²(c-3)- a²b(c-3)=
=(c-3)(30ab²-a²b)=
=ab(c-3)(30b-a)

ab(c-3)(30b-a) taeaz a=0,3,b=0,7 c=3i pięć siódmych
0,3×0,7(3i5/7-3)(30×0,7-0,3)=
=0,3×0,7(5/7)(21-0,3)=
=0,21×5/7×20,7=
=21/100×5/7×207/10=
=621/200=
=3i 21/200