Esce 4
Esce 4
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$\sqrt2(1-\sqrt3)+(\sqrt2-1)(\sqrt2+1)+\sqrt6=$
$=\sqrt2-\sqrt{2·3}+2\cancel{+\sqrt2}\cancel{-\sqrt2}-1+\sqrt6=$
$=\sqrt2\cancel{-\sqrt6}+2-1\cancel{+\sqrt6}=$
$=\sqrt2+1=$ riordina:
$=1+\sqrt2$
$(\approx{2,4142}).$
√2·(1 - √3) + (√2 - 1)·(√2 + 1) + √6=
=(√2 - √6) + 1 + √6=
=√2 + 1
ciao