Question

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Answer 1

$\displaystyle \Bigg(\frac{^{2-\sqrt{3}) }1}{2+\sqrt{3} } +\frac{^{\sqrt{3}+1) }2}{\sqrt{3} -1} \Bigg)\cdot \Bigg(\frac{^{3\sqrt{2}-4) }4}{3\sqrt{2} +4} -\frac{^{3+2\sqrt{2)} }3}{3-2\sqrt{2} } \Bigg)=\\\\=\Bigg(\frac{2-\sqrt{3} }{\Big(2-\sqrt{3} \Big)\Big(2+\sqrt{3}\Big) }+\frac{2\Big(\sqrt{3} +1\Big) }{\Big(\sqrt{3} -1\Big)\Big(\sqrt{3} +1\Big)} \Bigg) \cdot \Bigg(\frac{4\Big(3\sqrt{2}-4\Big) }{\Big(3\sqrt{2}-4\Big)\Big(3\sqrt{2}+4\Big) } -$

$\displaystyle -\frac{3\Big(3+2\sqrt{2}\Big) }{\Big(3-2\sqrt{2} \Big)\Big(3+2\sqrt{2} \Big)}\Bigg) =$

$\displaystyle =\Bigg(\frac{2-\sqrt{3} }{4-3} +\frac{2\sqrt{3}+2 }{3-1} \Bigg) \cdot \Bigg(\frac{12\sqrt{2}-16 }{18-16} -\frac{9+6\sqrt{2} }{9-8} \Bigg)=\\\\=\Bigg(\Big(2-\sqrt{3}\Big)+\frac{2\sqrt{3}+2 }{2} \Bigg)\cdot \Bigg(\frac{12\sqrt{2}-16 }{2} -\Big(9+6\sqrt{2}\Big)\Bigg)=$

$\displaystyle =\frac{2\Big(2-\sqrt{3}\Big) +2\sqrt{3} +2}{2} \cdot \frac{12\sqrt{2}-16-2\Big(9+6\sqrt{2}\Big) }{2} =\\\\ =\frac{4-2\sqrt{3}+2\sqrt{3} +2 }{2} \cdot \frac{12\sqrt{2}-16-18-12\sqrt{12} }{2} =\frac{6}{2} \cdot \Bigg(-\frac{34}{2} \Bigg)=\\\\=3 \cdot \Bigg(-\frac{34}{2} \Bigg)=-\frac{102}{2} =-51$

Answer 2

$\it Not\breve am\ a=\sqrt3 \Rightarrow a^2=3\ \ \ \ (1)\\ \\ Aducem\ la\ acela\d s\ numitor\ \hat \imath n\ prima \ parantez\breve a:\\ \\ \\ \dfrac{a-1+4+2a}{(2+a)(a-1)}=\dfrac{3a+3}{2a-2+a^2-a}=\dfrac{3a+3}{a-2+a^2}\ \stackrel{(1)}{=}\ \dfrac{3(a+1)}{a+1}=3$

$\it Not\breve am\ b=\sqrt2 \Rightarrow b^2=2\ \ \ \ (2)\\ \\ Aducem\ la\ acela\d s\ numitor\ \hat \imath n\ a\ doua\ a \ parantez\breve a:\\ \\ \\ \dfrac{12-8b-9b-12}{(3b+4)(3-2b)}=\dfrac{-17b}{9b-6b^2+12-8b}=\dfrac{-17b}{b-6b^2+12}\ \stackrel{(2)}{=}\ \dfrac{-17b}{b}=-17$

Acum,  expresia  dată devine:   3 · (-17)=- 51