сократите дробь 1) 5-√5/√10-5√2 4а^2+4а√б+б/4а^2-б 3) √3-3/3√2-√6 4) 9а^2-б^2/9а-6б√а+б^2 освободитесь от знака корня в знаменателе 1)10/3√5 2)11/2√3+1 3) 15/2√6 4)19/2√5-1

Ответы

gordeevnikita11 27.05.2020 07:26

$1) \frac{5-\sqrt{5}}{\sqrt{10}-5\sqrt{2}}=\frac{5-\sqrt{5}}{\sqrt{10}-\sqrt{50}}=\frac{(\sqrt{10}-\sqrt{50})(5-\sqrt{5})}{(\sqrt{10}+\sqrt{50})(\sqrt{10}-\sqrt{50})}=\frac{(\sqrt{10}-\sqrt{50})(5-\sqrt{5})}{10-50}=-\frac{10\sqrt{10}-30\sqrt{2}}{40}\\=0,75\sqrt{2}-0,25\sqrt{10}$

$2)\frac{ 4a^2+4a\sqrt{b}+b}{4a^2-b}=\frac{(2a+\sqrt{b})^2}{(2a+\sqrt{b})(2a-\sqrt{b})}=\frac{2a+\sqrt{b}}{2a-\sqrt{b}}$

$3) \frac{\sqrt{3}-3}{3\sqrt{2}-\sqrt{6}}= \frac{\sqrt{3}-3}{3\sqrt{2}-\sqrt{2}\sqrt{3}}= \frac{\sqrt{3}-3}{-\sqrt{2}(\sqrt{3}-3)}=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}$

Может опечатка ,а должно быть так:

$4)\frac{ 9a-b^2}{9a-6b\sqrt{a}+b^2}=\frac{ (3\sqrt{a}-b)(3\sqrt{a}+b)}{(3\sqrt{a}-b)^2}=\frac{ 3\sqrt{a}+b}{3\sqrt{a}-b}$

$1)\frac{10}{3\sqrt{5}}=\frac{10*\sqrt{5}}{3\sqrt{5}*\sqrt{5}}=\frac{10*\sqrt{5}}{15}$

$2)\frac{11}{2\sqrt{3}+1}=\frac{11(2\sqrt{3}-1)}{(2\sqrt{3}+1)(2\sqrt{3}-1)}=\frac{22\sqrt{3}-11}{(2\sqrt{3})^2-1^2}=\frac{22\sqrt{3}-11}{11}=2\sqrt{3}-1$

$3)\frac{15}{2\sqrt{6}}=\frac{15*\sqrt{6}}{2\sqrt{6}*\sqrt{6}}=\frac{15*\sqrt{6}}{2*6}=\frac{15*\sqrt{6}}{12}$

$4)\frac{19}{2\sqrt{5}-1}=\frac{19*(2\sqrt{5}+1)}{(2\sqrt{5}-1)*(2\sqrt{5}+1)}=\frac{38\sqrt{5}+19}{(2\sqrt{5})^2-1^2}=\frac{38\sqrt{5}+19}{19}=2\sqrt{5}+1$