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descompuneti in factori:
a)[tex] \sqrt{3 y^{2} } [/tex] + 2[tex] \sqrt{15}y[/tex] + 5[tex] \sqrt{3}[/tex]
b)[tex] {7} ay^{2} -28ay+28a[/tex]


Răspuns :

a) =rad3[rad(y^2)+2rad(5y)+5]=rad3(rady+rad5)^2

b)=7a(y^2-4a+4)=7a(y-2)^2

unde ^ inseamna putere