Răspuns:
Explicație pas cu pas:
[tex]x*y=x+y+x^{2} y^{2}[/tex]
a)
[tex]1*2=1+2+1^{2} *2^{2}=3+4=7[/tex]
b)
[tex]x*e=e*x=x => \\1)x*0=x+0+x^2*0^2=x\\2)0*x=0+x+0^2*x^2=x\\[/tex]
Din 1) si 2) => e=0 element neutru
c)
[tex](-2)*x\leq 3=> -2+x+(-2)^2*x^2\leq 3=>4x^2+x-2-3\leq 0=>4x^2+x-5\leq 0[/tex]
[tex]4x^2+x-5=0 \\d=b^2-4*a*c=1^2-4*4*(-5)=1+80=81\\x_{1}=\frac{-b+\sqrt{d}}{2*a}=\frac{-1+\sqrt{81}}{2*4}=\frac{-1+9}{8}=\frac{8}{8}=1\\x_{2}=\frac{-b-\sqrt{d}}{2*a}=\frac{-1-\sqrt{81}}{2*4}=\frac{-1-9}{8}=\frac{-10}{8}=\frac{-5}{4}[/tex]
x |-∞ [tex]\frac{-5}{4}[/tex] 1 +∞
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[tex]4x^{2}+x-5[/tex] |-∞ ++++++ 0 ------ 0 +++++++++++++ +∞
x∈[[tex]\frac{-5}{4}[/tex];1]
