Question

Помогите срочно нужно, на фото 2 вопроса, ДАЮ 50 БАЛОВ

Answer 1

$\displaystyle\bf\\6) \ \frac{\sqrt{54} -\sqrt{63} }{\sqrt{6}+ \sqrt{7} } -\frac{\sqrt{54} +\sqrt{63} }{\sqrt{6}- \sqrt{7} } =\\\\\\=\frac{\sqrt{9\cdot 6} -\sqrt{9\cdot 7} }{\sqrt{6}+ \sqrt{7} } -\frac{\sqrt{9\cdot 6} +\sqrt{9\cdot 7} }{\sqrt{6}- \sqrt{7} } =\\\\\\=\frac{3\sqrt{6} -3\sqrt{7} }{\sqrt{6}+ \sqrt{7} } -\frac{3\sqrt{6} +3\sqrt{7} }{\sqrt{6}- \sqrt{7} } =\\\\\\=\frac{3\cdot(\sqrt{6}-\sqrt{7})^{2}-3\cdot(\sqrt{6}+\sqrt{7})^{2} }{(\sqrt{6}+\sqrt{7} )\cdot(\sqrt{6}-\sqrt{7} ) } =$

$\displaystyle\bf\\=\frac{3\cdot(6-2\sqrt{42}+7-6-2\sqrt{42} -7) }{(\sqrt{6})^{2} -(\sqrt{7})^{2} } =\\\\\\=\frac{3\cdot(-4\sqrt{42} )}{6-7}=12\sqrt{42}$

Иррациональное

$\displaystyle\bf\\7) \ \sqrt{\frac{2-\sqrt{a} }{a+2\sqrt{a}+4 }:\frac{1}{8-a\sqrt{a} } }-\sqrt{a} =\\\\\\=\sqrt{\frac{2-\sqrt{a} }{\sqrt{a} +2\sqrt{a}+4 }:\frac{1}{2^{3} -(\sqrt{a})^{3} } }-\sqrt{a}=\\\\\\=\sqrt{\frac{(2-\sqrt{a})\cdot(2-\sqrt{a} )\cdot(a+2\sqrt{a}+4) }{a+2\sqrt{a}+4 } } }-\sqrt{a} =\\\\\\=\sqrt{(2-\sqrt{a} )^{2} } -\sqrt{a}=|2-\sqrt{a}|-\sqrt{a}=\\\\\\=\sqrt{a} -2-\sqrt{a} =-2$