Question

Punctele b) şi c).....

Answer 1

$b)x\circ x\circ x=[(x-4)^2+4]\circ x=[(x-4)^2+4-4](x-4)+4=\\(x-4)^3+4\\\text{Avem de rezolvat ecuatia:}\\(x-4)^3+4=x\\(x-4)^3+4-x=0\\(x-4)^3-(x-4)=0\\(x-4)[(x-4)^2-1]=0\\\text{Distingem doua cazuri diferite:}\\i)x-4=0\Rightarrow x=4\\ii)(x-4)^2-1=0\\(x-4)^2=1|\sqrt{} \\|x-4|=1\\ x-4=1\Rightarrow x=5\\x-4=-1\Rightarrow x=3\\S:x\in \{3,4,5\}$

$c)\text{De exemplu:}a=\dfrac{13}{5}\text{ si }b=\dfrac{38}{7} ,a\circ b=2\in \mathbb{N}$