Question

Прощу помощи помогите пж надо срочно

Answer 1

1)

а) $(2+\sqrt{3})*(1-\sqrt{3} ) = 2-2\sqrt{3} + \sqrt{3} -3 = -1-\sqrt{3}$

б) $(a+\sqrt{c})*(a-\sqrt{c}) = a^{2} -\sqrt{c}^{2} = a^{2} - c$

в) $(1+3\sqrt{2}) * (3\sqrt{2}  -1) = (3\sqrt{2}+1)*(3\sqrt{2}-1)=9*2-1=18-1=17$

г) $(1-2\sqrt{3})^{2}=1-4\sqrt{3}+12=13-4\sqrt{3}$

2)

а) $\frac{a^{2}-3 }{a+\sqrt{3} }=\frac{(a^{2}-3)*(a-\sqrt{3})}{(a+\sqrt{3})*(a-\sqrt{3})  } = \frac{(a^{2}-3) * (a-\sqrt{3})  }{a^{2}-3}=a-\sqrt{3}$

б) $\frac{\sqrt{7}-y }{7-y^{2} } = \frac{\sqrt{7} -y}{(\sqrt{7}-y) * (\sqrt{7} +y) } = \frac{1}{\sqrt{7} +y}$

в) $\frac{5-\sqrt{5} }{\sqrt{5} }=\frac{(5-\sqrt{5})\sqrt{5}  }{\sqrt{5} \sqrt{5} } = \frac{5\sqrt{5} -5}{5} = \frac{5(\sqrt{5}-1) }{5} = \sqrt{5} -1$

г) $\frac{\sqrt{b}-\sqrt{c}  }{b-c}= \frac{\sqrt{b}- \sqrt{c} }{(\sqrt{b} -\sqrt{c})*(\sqrt{b}  + \sqrt{c}) } = \frac{1}{\sqrt{b} +\sqrt{c} }$