Question

Ajutor ! Trebuie sa arăt că este patrat perfect .

Answer 1

[tex]\it 4+2\sqrt3 = 3+1+2\sqrt3=(\sqrt3)^2+2\sqrt3+1^2 =(\sqrt3+1)^2\ \ \ \ (*) \\\;\\ 4-2\sqrt3 = 3+1-2\sqrt3=(\sqrt3)^2-2\sqrt3+1^2 =(\sqrt3-1)^2[/tex]

[tex]\it \sqrt{(\sqrt3+1)^2} = |\sqrt3+1| = \sqrt3+1, deoarece\ \sqrt3+1\ \textgreater \ 0 \\\;\\ \sqrt{(\sqrt3-1)^2} = |\sqrt3-1| = \sqrt3-1, deoarece\ \sqrt3-1\ \textgreater \ 0[/tex]

Expresia din enunț devine:

$\it (\sqrt3+1+\sqrt3-1)^2+4= (2\sqrt3)^2 +4 = 12+4 = 16 = 4^2$

Answer 2

ridici la patrat cu curaj

4+2√3+4-2√3+2√(4²-(2√3)²)+4=
8+2√(16-12)+4=
8+2√4+4=
= 8+2*2+4=
=8+4+4=
=16=4², p.p.
as simple as that!!!