Question

$\frac{ \sqrt{6}- 2}{\sqrt{6}+2}: \frac{ \sqrt{3}- \sqrt{2} }{2}$   este nr nat.
{PS:a doua fracrie..V3-V2 totul la a doua doar partea de sus a fractiei}

Answer 1

Prima fractie o amplifici cu $\sqrt{6} -2$ si vei obtine: $\frac{ (\sqrt{6} -2)( \sqrt{6}-2) }{( \sqrt{6} +2)( \sqrt{6}-2) }= \frac{ ( \sqrt{6}-2) ^{2} }{6-4} = \frac{6-4 \sqrt{6}+4 }{2} = \frac{10- 4\sqrt{6} }{2}= \frac{2(5-2 \sqrt{6}) }{2}=5-2 \sqrt{6}$
Iar la a doua fractie: $\frac{( \sqrt{3} - \sqrt{2} )^2}{2} = \frac{3-2 \sqrt{6}+2 }{2}= \frac{5- 2\sqrt{6} }{2}$
Dupa ce faci calculele vei obtine ca: $\frac{5-2 \sqrt{6}}{\frac{5- 2\sqrt{6} }{2}} =2$