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July 2022 1 2 Report

№3 Сократить дробь
а)[tex]\frac{7+\sqrt{7} }{5\sqrt{7} }[/tex] б)[tex]\frac{16p-5}{4\sqrt{p}-\sqrt{5} }[/tex]
№4 Освободитесь от иррациональности в знаменателе
а)[tex]\frac{4}{2\sqrt{6} }[/tex] б)[tex]\frac{10}{3-\sqrt{5} }[/tex]

Answers & Comments

daraprelj

№3.а) $\displaystyle \frac{7+\sqrt{7} }{5\sqrt{7} } = \frac{\sqrt{7}*(\sqrt{7} +1) }{5\sqrt{7} }=\frac{\sqrt{7} +1 }{5 }$

№3.б) $\displaystyle \frac{16p-5}{4\sqrt{p}-\sqrt{5} } =\frac{(4\sqrt{p}+\sqrt{5} )(4\sqrt{p}-\sqrt{5} )}{4\sqrt{p}-\sqrt{5} } =4\sqrt{p}+\sqrt{5}$

№4.а) $\displaystyle \frac{4}{2\sqrt{6} } *\frac{\sqrt{6}}{\sqrt{6}} =\frac{4\sqrt{6}}{2*6} =\frac{\sqrt{6}}{3}$

№4.б) $\displaystyle \frac{10}{3-\sqrt{5} } *\frac{3+\sqrt{5}}{3+\sqrt{5}}=\frac{10*(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}=\frac{10*(3+\sqrt{5}) }{9-5} =\frac{10*(3+\sqrt{5}) }{4} =\frac{5*(3+\sqrt{5}) }{2} =\frac{15+5\sqrt{3} }{2}$

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