Odpowiedzi

2010-02-08T19:07:12+01:00
A)[2(x³-2x+4)-(x-3)(x²+5x-4)](3-2x)=
= [2x³-4x+8-(x³+5x²-4x-3x²-15x-12)](3-2x)=
= [2x³-4x+8-x³-5x²+4x+3x²+15x+12](3-2x)=
= [x³-2x²+15x+20](3-2x)=
=3x³-6x²+45x+60-2x⁴+4x³-30x-40x=
=-2x⁴+7x³-8x²-25x+60

b)[2(3x-4x+x³)-(2-x)(x²+6x-3)](2x-3)=
=[6x-8x+2x³-(2x²+12x-6-x³-6x²+3x)](2x-3)=
=[6x-8x+2x³-2x²-12x+6+x³+6x²-3x](2x-3)=
=[3x³+4x²-17x+6](2x-3)=
=6x⁴+8x³-34x²+12x-9x³-12x²+51x-16=
=6x⁴-x³-46x²+63x-16
2010-02-08T19:12:18+01:00
A) 2x(3)- 4x +8 - x(3) + 5x(2) - 4x - 3x(2) - 15x +12 (3-2x) =
x(3) -23x +20 +8x(2) (3-2x)=3x(3)-69 +60 +24x(2)-2x(4) -46x(2) -20x-16x(2)=3x(3) - 9 +6x(2) - 2x(4)
[to w nawiasie to liczba potegi np. 8x(2)]

b)6x-8x+2x(3) - 2x(2) +12x-6 - x(3) - 6x(2) + 3x (2x-3) =
10x +x(3)-8x(3) - 6 (2x-3)= 20x(2) +2x(4) - 16x(4)-12x -30x -3x(3) - 24x(3) - 18 = 20x(2)-14x(4) -42x - 27x(3)