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Bandarmariamagdalena
a fost răspuns

Se considera numere reale :a=[tex] \sqrt{3+2 \sqrt{2} } [/tex] si b=[tex] \sqrt{3-2 \sqrt{2} } [/tex]
a)Aratati ca a*b=1
b)Aratati ca [tex] a^{2} -2ab+ b^{2} =4
[/tex]
c)Demonstrati ca [tex] \frac{1}{a}- \frac{1}{b} [/tex]∈Q

Răspuns :

Renatemambouko
a). √(3+2√2) x √(3-2√2) =√[3²-(2√2)²] =√(9-8) = 1

b)  a²-2ab+b²=4
[√(3+2√2)]² - 2√(3+2√2) x√(3-2√2)+[√(3-2√2)]² =
3+2√2 - 2√(9-8)+3-2√2 = 6-2=4

c)  1/√(3+2√2) - 1/√(3-2√2) = [√(3-2√2) - √(3+2√2)]/(√9-8)=
=√(3-2√2) - √(3+2√2) /1 

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