Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 497

а) $$ \frac{\sqrt{70}-\sqrt{30}}{\sqrt{35}-\sqrt{15}}=\frac{\sqrt{2 \sqrt{35}}-\sqrt{2 \sqcap 5}}{\sqrt{35}-\sqrt{15}}= $$ $$ \frac{\sqrt{2} \sqrt{35}-\sqrt{2} \sqrt{15}}{\sqrt{35}-\sqrt{15}}=\frac{\sqrt{2}(\sqrt{35}-\sqrt{15})}{\sqrt{35}-\sqrt{15}}= $$ $$ \frac{\sqrt{2}(\sqrt{35}-\sqrt{15}) \div(\sqrt{35}-\sqrt{15})}{(\sqrt{35}-\sqrt{15}) \div(\sqrt{35}-\sqrt{15})} $$ = $$\sqrt{2}$$
б) $$ \frac{\sqrt{15}-5}{\sqrt{6}-\sqrt{10}}=\frac{\sqrt{53}-(\sqrt{5})^2}{\sqrt{23}-\sqrt{25}}= $$ $$ \frac{\sqrt{5} \square \sqrt{3}-\sqrt{5} \square \sqrt{5}}{\sqrt{2} \square \sqrt{3}-\sqrt{2} \sqrt{5}}=\frac{\sqrt{5}(\sqrt{3}-\sqrt{5})}{\sqrt{2}(\sqrt{3}-\sqrt{5})}= $$ $$ \frac{\sqrt{5}(\sqrt{3}-\sqrt{5}) \div(\sqrt{3}-\sqrt{5})}{\sqrt{2}(\sqrt{3}-\sqrt{5}) \div(\sqrt{3}-\sqrt{5})}= $$ $$ \frac{\sqrt{5}}{\sqrt{2}}=\frac{\sqrt{5} \square \sqrt{2}}{\sqrt{2} \square \sqrt{2}}=\frac{\sqrt{10}}{2} $$
в) $$ \frac{9-2 \sqrt{3}}{3 \sqrt{6}-2 \sqrt{2}}=\frac{3(\sqrt{3})^2-2 \sqrt{3}}{3 \sqrt{32}-2 \sqrt{2}}= $$ $$ \frac{\sqrt{3}(3 \sqrt{3}-2)}{\sqrt{2}(3 \sqrt{3}-2)}=\frac{\sqrt{3}(3 \sqrt{3}-2) \div(3 \sqrt{3}-2)}{\sqrt{2}(3 \sqrt{3}-2) \div(3 \sqrt{3}-2)} $$ $$ =\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{3} \sqrt{2}}{\sqrt{2} \sqrt{2}}=\frac{\sqrt{6}}{2} $$
г) $$ \frac{2 \sqrt{3}+3 \sqrt{2}-\sqrt{6}}{2+\sqrt{6}-\sqrt{2}}=\frac{2 \sqrt{3}+(\sqrt{3})^2 \sqrt{2}-\sqrt{32}}{2+\sqrt{6}-\sqrt{2}}= $$ $$ \frac{\sqrt{3}(2+\sqrt{3}-\sqrt{2}-\sqrt{2})}{2+\sqrt{6}-\sqrt{2}}=\frac{\sqrt{3}(2+\sqrt{6}-\sqrt{2})}{2+\sqrt{6}-\sqrt{2}}= $$ $$ \frac{\sqrt{3}(2+\sqrt{6}-\sqrt{2}) \div(2+\sqrt{6}-\sqrt{2})}{(2+\sqrt{6}-\sqrt{2}) \div(2+\sqrt{6}-\sqrt{2})} $$ = $$\sqrt{3}$$