Odpowiedzi

2010-02-26T12:31:51+01:00
1)
-x³+2x²-x+5-[2x²-x+1+3(5x+2)]=-x³+2x²-x+5 -[2x²-x+1+15x+6]=-x³+2x²-x+5-2x²+x-1-15-6=-x³-15x-1
2)
2(-x³+2x²-x+5)+3[5x+2-2(2x²-x+1)=-2x³+4x²-2x+10+3[5x+2-4x²+2x-2]=-2x³+4x²-2x+10+15x+6-12x²+6x-6=-2x³-8x²+19x+10
3)
5(5x+2)-[-x³+2x²-x+5+2(2x²-x+1)]=25x+10-[-x³+2x²-x+5+4x²-2x+2]=25x+10+x³-2x²+x-5-4x²+2x-2=x³-6x²+28x+3
4)
3[-x³+2x²-x+5-2(2x²-x+1)]-[2x²-x+1+3(-x³+2x²-x+5)]=3[-x³+2x²-x+5-4x²+2x-2]-[2x²-x+1-3x³+6x²-3x+15]=-3x³+6x²-3x+15-12x²+6x-6-2x²+x-1+3x³-6x²+3x-15=-14x²+7x-7