Question

√3-√2 supra lui √3+√2 rog ajutor​

Answer 1

Răspuns:

$5-2\sqrt{6}$; $\frac{6\sqrt{3}-\sqrt{15} }{18}$

Explicație pas cu pas:

$\frac{\sqrt{3}-\sqrt{2} }{\sqrt{3}+\sqrt{2} } =\frac{(\sqrt{3}-\sqrt{2} )(\sqrt{3}-\sqrt{2} )}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2} )} =\frac{(\sqrt{3}-\sqrt{2} )^{2} }{3-2}=5-2\sqrt{6}$

$\frac{6-\sqrt{5} }{3\sqrt{12} } =\frac{6-\sqrt{5} }{3\sqrt{2^{2}*3 } } =\frac{6-\sqrt{5} }{6\sqrt{3} }=\frac{(6-\sqrt{5})\sqrt{3} }{6\sqrt{3}\sqrt{3} } =\frac{6\sqrt{3}-\sqrt{15} }{18}$