Question

vreau si eu la b si c va rog

Answer 1

[tex]b)x\ast y=4(x+y-3)-xy=4x+4y-12-xy=\\ =4 +4x+4y-xy-16 =x(4-y)-4(4-y) +4= \\ =(4-y)(x-4)+4= 4-(x-4)(y-4)\\ \\ c)x\ast x=4(x+x-3)-x^2= 8x-12-x^2\\ x\ast x\ast x=4(x+8x-12-x^2-3)-x(8x-12-x^2)\\ x\ast x\ast x=4(9x-x^2-15)-8x^2+12x+x^3\\ x\ast x\ast x=36x-4x^2-60-8x^2+12x+x^3\\ x\ast x\ast x=x^3-12x^2+48x-60 \\ \text{Avem de rezolvat ecuatia:}\\ x^3-12x^2+48x-60=x\\ [/tex]
[tex]x^3-12x^2+47x-60=0\\ \text{Ramane sa descompunem ecuatia.Sa observam ca 3 este solutie}\\ x^3-3x^2-9x^2+27x+20x-60=0\\ x^2(x-3)-9x(x-3)+20(x-3)=0\\ (x-3)(x^2-9x+20)=0\\ \text{Care se descompune mai departe:}\\ (x-3)(x-4)(x-5)=0\\ S=\{3,4,5\} [/tex]