Răspuns :
[tex]x * y=x y-\sqrt{2} x-\sqrt{2} y+\sqrt{2}+2[/tex]
1)
√2*(-√2)=√2
Il inlocuim pe x cu √2 si pe y cu -√2 si obtinem:
-2-2+2+√2+2=√2
2)
[tex]x y-\sqrt{2} x-\sqrt{2} y+\sqrt{2}+2=x y-\sqrt{2} x-\sqrt{2} y+2+\sqrt{2}=x(y-\sqrt{2})-\sqrt{2}(y-2)+\sqrt{2} =(y-\sqrt{2})(x-\sqrt{2})+\sqrt{2}[/tex]
3)
Elementul neutru
x*e=x
[tex]x e-\sqrt{2} x-\sqrt{2} e+\sqrt{2}+2=x\\\\(x-\sqrt{2})(e-\sqrt{2})+\sqrt{2} =x\\\\ (x-\sqrt{2})(e-\sqrt{2})-(x-\sqrt{2})=0\\\\ (x-\sqrt{2})(e-\sqrt{2}-1)=0 \\\\e-\sqrt{2}-1=0\\\\e=1+\sqrt{2}[/tex]
4)
a*a=2+√2
[tex](a-\sqrt{2}) (a-\sqrt{2}) +\sqrt{2}=2+\sqrt{2} \\\\(a-\sqrt{2})^2=2\\\\ a-\sqrt{2}=-\sqrt{2}\\\\ a=0\\\\ \\a-\sqrt{2}=\sqrt{2}\\\\ a=2\sqrt{2}[/tex]
5)
[tex]4^x*2^x=\sqrt{2}\\\\ (4^x-\sqrt{2})(2^x-\sqrt{2})+\sqrt{2}=\sqrt{2}\\\\ (4^x-\sqrt{2})(2^x-\sqrt{2})=0\\\\4^x-\sqrt{2}=0\\\\4^x=2^{\frac{1}{2} }\\\\2x=\frac{1}{2}\\\\ x=\frac{1}{4} \\\\\\2^x-\sqrt{2}=0\\\\2^x=\sqrt{2} \\\\x=\frac{1}{2}[/tex]
6)
[tex](x+\sqrt{2} )*(x-\sqrt{2})\leq \sqrt{2}\\\\x\cdot (x-2\sqrt{2})+\sqrt{2}\leq \sqrt{2}\\\\x\cdot (x-2\sqrt{2})\leq 0[/tex]
x=0 si x=2√2
Facem tabel semn
x -∞ 0 2√2 +∞
x(x-2√2) + + + + + 0- - - - 0+ + + ++
x∈[0,2√2]
Un alt exercitiu cu legi de compozitie gasesti aici: https://brainly.ro/tema/4728041
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