Question

Rozwiąż::
A)(x-2)(2X+3)(X²-4)=0
B)X³+X²-X+1=0
C)X³-13X+12=0
D)(X-1)(X+2)(x+2)(x-3)≤0
E)(X²-1)(x²-4)>0
f)x³-2x²+4x-8≤0

Answer 1

A)(x-2)(2X+3)(X²-4)=0
(2x²+3x-4x-6)(x²-4)=0
2x(4)-8x²+3x³-12x-4x³+16x-6x²+24=0
2x(4)-x³-14x²+4x=-24
2x(4)- tak zapisalam 2x do potegi4

B)X³+X²-X+1=0
x³+x²-x=-1

C)X³-13X+12=0
x³-13x=-12
D)(X-1)(X+2)(x+2)(x-3)≤0
(x²+2x-x-2)(x²-3x+2x-6)≤0
(x²+x-2)(x²-x-6)≤0
x(4)-x³-6x²+x³-x²-6x-2x²+2x+12≤0
x(4)-9x²-4x≤-12

E)(X²-1)(x²-4)>0
x(4)-4x²-x² >-4

f)x³-2x²+4x-8≤0
x³-2x²+4x≤8