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Question 1

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Question 2

А) $\frac{ x^{2} +(2- \sqrt{3} )x-2 \sqrt{3} }{ x^{2} -( \sqrt{3}+1)x+ \sqrt{3}}= \frac{ x^{2} +2x- x\sqrt{3}-2 \sqrt{3} }{ x^{2} -x \sqrt{3}-x+ \sqrt{3}}= \frac{x(x+2)-\sqrt{3}(x+2)}{x(x-1)-\sqrt{3}(x-1)}= \frac{(x-\sqrt{3})(x+2)}{(x-1)(x-\sqrt{3})}$
$= \frac{x+2}{x-1}$
б) $\frac{ a^{2} +( \sqrt{2}-1)a-\sqrt{2}}{ \sqrt{2}+(1-3 \sqrt{2})a-3 a^{2}}=\frac{ a^{2} +a \sqrt{2}-a-\sqrt{2}}{\sqrt{2}+a-3a\sqrt{2}-3a^{2}}=\frac{ a(a+\sqrt{2})-(a+\sqrt{2})}{(\sqrt{2}+a)-3a( \sqrt{2}+a)}=\frac{(a+\sqrt{2})(a-1)}{(\sqrt{2}+a)(1-3a)}$ $= \frac{a-1}{1-3a}$
в) $\frac{ b^{2} -( \sqrt{2} - \sqrt{3})b- \sqrt{6} }{- b^{2}+( \sqrt{5} - \sqrt{3} )b+ \sqrt{15} } = \frac{ b^{2} -b \sqrt{2} +b \sqrt{3}- \sqrt{6} }{- b^{2}+b\sqrt{5} -b \sqrt{3}+ \sqrt{15} } = \frac{ b(b- \sqrt{2})+\sqrt{3}(b-\sqrt{2}) }{-b(b+\sqrt{3} )+ \sqrt{5}(b+ \sqrt{3} )} =$
$= \frac{(b- \sqrt{2} )(b+ \sqrt{3} )}{(b+ \sqrt{3} )( \sqrt{5} -b)} = \frac{b- \sqrt{2} }{ \sqrt{5} -b}$
г) $\frac{3 y^{2} +(3 \sqrt{6}- \sqrt{2} )y-2 \sqrt{3} }{2 y^{2}-( \sqrt{3}-2 \sqrt{6} )y-3 \sqrt{2}} = \frac{3 y^{2} +3y \sqrt{6}-y \sqrt{2}-2 \sqrt{3} }{2 y^{2}-y \sqrt{3}+2y \sqrt{6}-3 \sqrt{2}} = \frac{(3 y^{2}-y \sqrt{2})+(3y \sqrt{6}-2 \sqrt{3} )}{(2 y^{2}-y \sqrt{3})+(2y \sqrt{6}-3 \sqrt{2})} =$
$= \frac{y(3 y-\sqrt{2})+(3y \sqrt{2}-2) \sqrt{3}}{y(2 y-\sqrt{3})+(2y \sqrt{3}-3) \sqrt{2}} = \frac{y(3 y-\sqrt{2})+(3y- \sqrt{2}) \sqrt{3} \sqrt{2} }{y(2 y-\sqrt{3})+(2y -\sqrt{3}) \sqrt{2} \sqrt{3} } = \frac{(y+ \sqrt{6} )(3y- \sqrt{2})}{(y+ \sqrt{6} )(2y -\sqrt{3})} =$
$= \frac{3y- \sqrt{2}}{2y -\sqrt{3}}$