Question

Bunaa!! Am nevoie de ajutor la cele doua exercitii!! Dau coronita !!

Answer 1

a) x=$\sqrt{288} = 12\sqrt{2}$    ( am scos radicalul ; 288=$12^{2} x2$)
y =$\sqrt{162} = 9\sqrt{2}$                ( 162=$9^{2} x2$ )
media geometrica =$\sqrt{xy}$
dar xy=$12 \sqrt{2} x 9 \sqrt{2} =12x9x2=216$
asadar , media geometrica = $\sqrt{216}$=$6 \sqrt{6}$

b)-3x+1=7
   -3x=6
     x= - 2  ∈ R