Bonsoir,On pose A = 1-√2 et B = [tex]\frac{3-\sqrt{18} }{3-2\sqrt{2} }[/tex]Calculer A², puis montre.... Pergunta de ideia deMartin1776

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Bonsoir,

On pose A = 1-√2 et B = [tex]\frac{3-\sqrt{18} }{3-2\sqrt{2} }[/tex]

Calculer A², puis montrer que B = -3-3√2

Merci beaucoup

hatisham

A=1-√2

A²=(1-√2)²

=1-2√2+2

=3+2√2

B= $\frac{3-\sqrt{18} }{3-2\sqrt{2} } =\frac{(3-\sqrt{18})(3+2\sqrt{2}) }{(3-2\sqrt{2})(3+2\sqrt{2} ) }$

= $\frac{9+6\sqrt{2}-3\sqrt{18}-2\sqrt{2*18} }{9-8} =\frac{9+6\sqrt{2}-3*3\sqrt{2}-2*3\sqrt{2*2} }{1}$ car √18=3√2

=9+6√2-9√2-6*2

=9-12+6√2-9√2

= -3-3√2

* veut dire multiplication

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