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Nicageorgianama
a fost răspuns

{x∈NΙx^2+x-2=0} ={1}
{x∈ZΙ2x^2-3x+1=0} ={1}

Răspuns :

Lozanalex
{x∈NΙx^2+x-2=0} ={1}
x*x+x-2=0
(delta) d=1*1-4*1*(-2)=1+8=9
x1=(-1-3)/(2*1)=-4/2=-2
x2=(-1+3)/(2*1)=2/2=1

{x∈ZΙ2x^2-3x+1=0} ={1}
2*x*x-3*x+1=0
d=(-3)*(-3)-4*2*1=9-8=1
x1=(3-1)/(2*2)=2/4=1/2=0.5
x2=(3+1)/(2*2)=4/4=1

Cea ce trebuia de demonstrat

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