Question

Aratati ca nr : A=(2-radical din 3)^2+(3 +radical din 3)^2 - 2 radical din 3 este natural

Urgent! Va roog !

Answer 1

Răspuns:

A = 19 nr natural

Explicație pas cu pas:

Formule

(a+b)^2 = a^2 + 2ab + b^2

(a-b)^2 = a^2 -2ab + b^2

Deci

A = 4 - 4radical(3) + 3 + 9 + 6radical(3) + 3 - 2radical(3)

= 4 + 3 + 3 + 9 - 6radical(3) + 6radical(3) = 19

Answer 2

$A=(2-\sqrt{3} )^{2} +(3+\sqrt{3} )^{2}-2\sqrt{3} \\A=(2^{2} -2*2*\sqrt{3} +(-\sqrt{3} )^{2})+(3^{2}+2*3*\sqrt{3} +\sqrt{3} ^{2})-2\sqrt{3}\\ A=(4-4\sqrt{3} +3)+(9+6\sqrt{3}+3)-2\sqrt{3}\\A=7-4\sqrt{3}+12+6\sqrt{3}-2\sqrt{3}\\A=19+2\sqrt{3}-2\sqrt{3}\\A=19$

=>A∈N