Решите любые номера но не меньше 3 и с решениями ща старания 30 .

Ответ:

elizavetaliza261204

17.08.2020 07:30

21.

$\frac{2}{5\sqrt{3} } =\frac{2*\sqrt{3} }{5*\sqrt{3} *\sqrt{3} } =\frac{2\sqrt{3} }{5*3} =\frac{2\sqrt{3} }{15} \\ \\\frac{\sqrt{5}+2 }{\sqrt{5} -2} =\frac{(\sqrt{5}+2)(\sqrt{5}+2)}{(\sqrt{5}-2)(\sqrt{5}+2)} =\frac{5+4\sqrt{5}+4 }{5-4} =\frac{9+4\sqrt{5} }{3} \\ \\ \frac{17}{3\sqrt{2}+1 } =\frac{17(3\sqrt{2}-1)}{(3\sqrt{2}+1)(3\sqrt{2}-1)}= \frac{17(3\sqrt{2}-1)}{9*2-1} =\frac{17(3\sqrt{2}-1)}{17} =3\sqrt{2} -1\\ \\$ $\\ \frac{4}{2\sqrt{5} +\sqrt{12} } =\frac{4(2\sqrt{5} -\sqrt{12} )}{(2\sqrt{5} +\sqrt{12})(2\sqrt{5} -\sqrt{12} )}}=\frac{4(2\sqrt{5} -\sqrt{12} )}{4*5-12} =\frac{4(2\sqrt{5} -\sqrt{12} )}{8} =\frac{2\sqrt{5} -\sqrt{12}}{2}$

$\\ \frac{7}{\sqrt{18}+2\sqrt{2} +1 } =\frac{7}{3\sqrt{2}+2\sqrt{2} +1 } =\frac{7}{5\sqrt{2}+1 } =\frac{7(5\sqrt{2}-1)}{(5\sqrt{2}+1)(5\sqrt{2}-1)} =\frac{7(5\sqrt{2}-1)}{25*2-1}\\ \\ =\frac{7(5\sqrt{2}-1)}{49} =\frac{5\sqrt{2}-1}{7}$

22.

$a=\sqrt{5} -\frac{1}{\sqrt{5}-2} =\sqrt{5} -\frac{\sqrt{5} +2}{(\sqrt{5}-2)(\sqrt{5} +2) } =\sqrt{5} -\frac{\sqrt{5} +2}{5-4} =\sqrt{5} -\sqrt{5} +2=2$ - выражение рациональное

$b=\sqrt{3} -\frac{1}{2+\sqrt{3} } =\sqrt{3} -\frac{2-\sqrt{3} }{(2+\sqrt{3} )(2-\sqrt{3} )} =\sqrt{3} -\frac{2-\sqrt{3} }{4-3} =\sqrt{3}-2+\sqrt{3} =2\sqrt{3} -2$ - иррациональное

$c=\sqrt{5}+\frac{1}{\sqrt{5} +2}=\sqrt{5} +\frac{\sqrt{5} -2}{(\sqrt{5} +2)(\sqrt{5} -2)} =\sqrt{5} +\frac{\sqrt{5} -2}{5-4} =\sqrt{5} +\sqrt{5} -2=2\sqrt{5} -2$ - иррациональное

25.

$\frac{4+\sqrt{3} }{2+\sqrt{3}} +\frac{4-\sqrt{3} }{2-\sqrt{3}} =\frac{(4+\sqrt{3})(2-\sqrt{3}) +(4-\sqrt{3})(2+\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} =\frac{8-4\sqrt{3}+2\sqrt{3}-3+8+4\sqrt{3}-2\sqrt{3} -3}{4-3}=10\\ \\ \\ \frac{3+\sqrt{7} }{3-\sqrt{7} } +\frac{3-\sqrt{7} }{3+\sqrt{7}}=\frac{(3+\sqrt{7})(3-\sqrt{7})+(3-\sqrt{7})(3+\sqrt{7})}{(3-\sqrt{7})(3+\sqrt{7})}=\frac{2(3-\sqrt{7})(3+\sqrt{7})}{(3-\sqrt{7})(3+\sqrt{7})} =2$

$\frac{\sqrt{3} }{\sqrt{3} -\sqrt{2} } -\frac{\sqrt{2} }{\sqrt{3} +\sqrt{2} } =\frac{\sqrt{3}(\sqrt{3} +\sqrt{2} )-\sqrt{2}( \sqrt{3} -\sqrt{2} ) }{(\sqrt{3} -\sqrt{2} )(\sqrt{3} +\sqrt{2} )} =\frac{3+\sqrt{6} -\sqrt{6} +2}{3-2} =3$