[tex]x * y=5(x+2)(y+2)-2[/tex]
a)
Inlocuim pe y cu -2 si obtinem:
x*(-2)=-2
5(x+2)(-2+2)-2=5(x+2)×0-2=0-2=-2
b)
[tex]f(x+y)=\frac{e^{x+y}-10}{5} \\\\f(x)*f(y)=5(\frac{e^x-10}{5} +2)(\frac{e^y-10}{5} +2)-2=e^x(\frac{e^y-10}{5} +2)-2=e^x\cdot \frac{e^y}{5} -2=\frac{e^{x+y}-10}{5}=f(x+y)[/tex]
c)
x*x=5(x+2)²-2
x*x*x=[5(x+2)²-2]*x=5(5(x+2)²-2+2)(x+2)-2=25(x+2)³-2
25(x+2)³-2=23
25(x+2)³=25
(x+2)³=1
x+2=1
x=-1
Un alt exercitiu cu legi de compozitie gasesti aici: https://brainly.ro/tema/9918936
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