Question

докажите тождество
$\frac{37}{7 + 2 \sqrt{3} } + \frac{37}{7 - 2 \sqrt{3} } = 14$
$\frac{ \sqrt{10} + 3 }{ \sqrt{10} - 3} + \frac{ \sqrt{10} - 3}{ \sqrt{10 } + 3 } = 38$
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Answer 1

(_/10+3)*(_/10+3) +(_/10-3)*(_/10-3)=(_/10+3)^2+ (_/10-3)^2= 10+ 6_/10+9 + 10- 6_/10+ 9=10+10+9+9=38
38=38

Answer 2

$\frac{37}{7+2\sqrt3}+\frac{37}{7-2\sqrt3}=\frac{37(7-2\sqrt3)}{49-4\cdot 3}+\frac{37(7+2\sqrt3)}{49-4\cdot 3}=\frac{259-74\sqrt3+259-74\sqrt3}{37}=\\\\=\frac{518}{37}=14\\\\\\\frac{\sqrt{10}+3}{\sqrt{10}-3}+\frac{\sqrt{10}-3}{\sqrt{10}+3}=\frac{(\sqrt{10}+3)^2}{(\sqrt{10}-3)(\sqrt{10}+3)}+\frac{(\sqrt{10}-3)^2}{(\sqrt{10}-3)(\sqrt{10}+3)}=\\\\=\frac{10+6\sqrt{10}+9+10-6\sqrt{10}+9}{(\sqrt{10}-3)(\sqrt{10}+3)}=\frac{2\cdot 19}{10-9}=38$