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Daca a,b,c aprtine de Q si ab +ac+bc=100 aratati ca numarul
x=radical din (100+aLa a doua)+(100+b la a doua )+(100+c la a doua )
este numar rational
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Răspuns :

[tex]\displaystyle\it\\ab+ac+bc=100,~\sqrt{(100+a^2)(100+b^2)(100+c^2)}\in\mathbb{Q}~?\\a^2+100=ab+ac+bc+a^2=b(a+c)+a(a+c)=(a+b)(a+c).\\b^2+100=ab+ac+bc+b^2=(b+c)(a+b).\\c^2+100=ab+ac+bc+c^2=(b+c)(a+c).\\deci,~\sqrt{(100+a^2)(100+b^2)(100+c^2)}=\\\sqrt{(a+b)(a+c)(b+c)(a+b)(b+c)(a+c)}=\sqrt{(a+b)^2(a+c)^2(b+c)^2}=\\(a+b)(a+c)(b+c) \in\mathbb{Q}~pentru~ca~si~a,b,c\in\mathbb{Q}.[/tex]

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