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[tex]Fie~a,b,c\in Q~si~a \sqrt{2} +b \sqrt{3} +c \sqrt{5} =0 \\ Sa~se~arate~ca~a=b=c=0. \\ Rezolvare~: \\ a \sqrt{2} +b \sqrt{3} +c \sqrt{5} =0~|* \sqrt{2} \\ a+ b\sqrt{6} +c \sqrt{10} =0 \\ a=\underbrace{-( b\sqrt{6} +c \sqrt{10} )}_{\in R\backslash{Q}} \\ \\ a=0 \\ Daca~a=0\Longrightarrow b \sqrt{3} +c \sqrt{5} =0~|* \sqrt{3} \\ b+c \sqrt{15} =0 \\ b=-c \sqrt{15} \Longrightarrow~b=0 \\ c~ \sqrt{5} =0 \Longrightarrow~c=0 \\ Deci~a=b=c=0 \\ \\ E ~bine?[/tex]

Răspuns :

[tex]\displaystyle a \sqrt{2}+b \sqrt{3}+c \sqrt{5}=0 \\ \\ Avem~a \sqrt{2}+b \sqrt{3}=-c \sqrt{5}.~Ridicand~la~patrat,~obtinem: \\ \\ 2a^2+2ab \sqrt{6}+3b^2=5c^2. \\ \\ Rezulta~2ab \sqrt{6} \in \mathbb{Q} \Rightarrow ab=0. \\ \\ \bullet Daca~a=0 \Rightarrow b \sqrt{3}+ c \sqrt{5}=0.~Rezulta~3b^2+2bc \sqrt{15}+5c^2=0. \\ \\ Deci~2bc \sqrt{15} \in \mathbb{Q} \Rightarrow bc=0. \\ \\ -Daca~b=0 \Rightarrow c=0. \\ \\ - Daca~c=0 \Rightarrow b=0. \\ \\ Deci~in~acest~caz~avem~a=b=c=0.[/tex]

[tex]\displaystyle \bullet Daca~b=0 \Rightarrow a \sqrt{2}+c \sqrt{5}=0.~Deci~2a^2+2ac \sqrt{10}+5c^2=0. \\ \\ Deci~2ac \sqrt{10} \in \mathbb{Q} \Rightarrow ac=0. \\ \\ - Daca~a=0 \Rightarrow c=0. \\ \\ -Daca~c=0 \Rightarrow a=0. \\ \\ Am~obtinut~din~nou~solutia~a=b=c=0.[/tex]