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


Răspuns :

[√(3+2√2) - √(3-2√2) ]² =
=3+2√2 -2(√9-8)+3-2√2=
=3 -2*1+3=4

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