Question

Fie numerele a= \sqrt{2- \sqrt{2} } si b= \sqrt{2+ \sqrt{2} } .
A) Calculati axb.
B) Calculati (a+b)^{2}.
C) Aratati ca numarul b:a- \sqrt{2} este rational.

(P.S.: Cine raspunde la toate subpunctele corect primeste coroana)

Answer 1

A)

[tex]ab = \sqrt{2- \sqrt{2} } * \sqrt{2+ \sqrt{2} } = \sqrt{(2- \sqrt{2})(2+ \sqrt{2}) } =\\ = \sqrt{2 ^{2} - \sqrt{2} ^{2} } = \sqrt{2} \\[/tex]

B)

$(a+b)^{2} = a^{2} + b^{2} + 2ab = (2- \sqrt{2}) + (2+ \sqrt{2}) + 2 \sqrt{2} =\\ =4 + 2 \sqrt{2}$

Answer 2

a·b=√2

(a+b)² = a² +2ab +b²


[tex]\dfrac{a}{b} = \sqrt{\dfrac{2-\sqrt2}{2+\sqrt2}} =\sqrt{\dfrac{(2+\sqrt2)^2}{2}} = \dfrac{2+\sqrt2}{\sqrt2} =\dfrac{\sqrt2(\sqrt2+1)}{\sqrt2} =\sqrt2+1 \\\;\\ \dfrac{a}{b} -\sqrt2 =\sqrt2+1-\sqrt2 =1 \in \mathbb{Q}[/tex]