Question

Дуже потрібна допомога ​

Answer 1

$\displaystyle\bf\\1)\\\\\frac{40}{8-\sqrt{24} } =\frac{40\cdot(8+\sqrt{24})}{(8-\sqrt{24})\cdot(8+\sqrt{24} ) } =\frac{40\cdot(8+\sqrt{24})}{8^{2} -(\sqrt{24})^{2} } =\\\\\\=\frac{40\cdot(8+\sqrt{24})}{64-24 } =\frac{40\cdot(8+\sqrt{24})}{40 }=8+\sqrt{24} \\\\2)\\\\\frac{10}{\sqrt{34} +\sqrt{24} } =\frac{10\cdot(\sqrt{34} -\sqrt{24})}{(\sqrt{34} +\sqrt{24})\cdot(\sqrt{34} -\sqrt{24} ) } =\frac{10\cdot(\sqrt{34} -\sqrt{24})}{(\sqrt{34})^{2} -(\sqrt{24})^{2} } =$

$\displaystyle\bf\\=\frac{10\cdot(\sqrt{34} -\sqrt{24})}{34-24 } =\frac{10\cdot(\sqrt{34} -\sqrt{24})}{10 }=\sqrt{34}- \sqrt{24} \\\\3)\\\\\frac{110}{12+\sqrt{34} } =\frac{110\cdot(12-\sqrt{34}) }{(12+\sqrt{34})\cdot(12-\sqrt{34}) } =\frac{110\cdot(12-\sqrt{34}) }{12^{2} -(\sqrt{34})^{2} }=\\\\\\=\frac{110\cdot(12-\sqrt{34}) }{144-34 }=\frac{110\cdot(12-\sqrt{34}) }{110 }=12-\sqrt{34} \\\\4)$

$\displaystyle\bf\\\frac{40}{8-\sqrt{24} } +\frac{10}{\sqrt{34} +\sqrt{24} }+ \frac{110}{12+\sqrt{34} } =\\\\\\=8+\sqrt{24}+\sqrt{34} -\sqrt{24} +12-\sqrt{34} =8+12=20\\\\\\Otvet \ : \ 20$