Задания простенькие.Ниже в коментариях(ссылка) на еще 2 такое же задания. Решить не просто дав ответ----- больше ПРИМЕР ((Корень 49*81=63)) А решить нормально то есть вот так------ больше ПРИМЕР ((Корень 49*81=49*81=корень 3969...

[latex]1)\; \; 8\sqrt{0,1}-2\sqrt{0,4}-0,5\sqrt{40}=8\sqrt{0,1}-2\sqrt{4\cdot 0,1}-0,5\sqrt{400\cdot 0,1}=\\\\=8\sqrt{0,1}-2\cdot 2\sqrt{0,1}-0,5\cdot 20\sqrt{0,1}=8\sqrt{0,1}-4\sqrt{0,1}-10\sqrt{0,1}=\\\\=-6\sqrt{0,1}\\\\2)\; \; \sqrt{ \frac{2}{3}} + \sqrt{ \frac{50}{3}}-\sqrt6= \sqrt{ \frac{2}{3} } + \sqrt{ \frac{25\cdot 2}{3} } -\sqrt6= \sqrt{ \frac{2}{3} } +5 \sqrt{ \frac{2}{3} } -\sqrt{6}=[/latex] [latex]=6\sqrt{\frac{2}{3}}-\sqrt6=6\cdot \frac{\sqrt6}{3}-\sqrt6=2\sqrt6-\sqrt6=\sqrt6[/latex] [latex]3)\; \; \sqrt{10}-\sqrt{\frac{605}{8}}+\frac{99}{4}\sqrt{\frac{10}{121}}=\sqrt{10}-\frac{\sqrt{5\cdot 121}}{\sqrt{2^3}}+ \frac{99}{4}\cdot \frac{\sqrt{10}}{\sqrt{11^2}} =\\\\=\sqrt{10}- \frac{11\sqrt5}{2\sqrt2}+\frac{99}{4}\cdot \frac{\sqrt{10}}{11}=\sqrt{10}-\frac{11}{2}\cdot \frac{\sqrt{5}}{\sqrt2}+\frac{9\cdot 11\sqrt{10}}{4\cdot 11} =[/latex] [latex]= \sqrt{10} -\frac{11}{2} \cdot \frac{\sqrt5\cdot \sqrt2}{\sqrt2\cdot \sqrt2}+\frac{9}{4}\cdot \sqrt{10}=\sqrt{10}-\frac{11}{2}\cdot \frac{\sqrt{10}}{2}+\frac{9}{4}\cdot \sqrt{10}=[/latex] [latex]=\sqrt{10}-\frac{11}{4}\cdor \sqrt{10}+\frac{9}{4}\cdot \sqrt{10}=\sqrt{10}\cdot (1-\frac{11}{4}+\frac{9}{4})=\\\\=\sqrt{10}\cdot \frac{1}{2}=\frac{\sqrt2\cdot \sqrt5}{2}= \frac{\sqrt5}{\sqrt2} =\sqrt{\frac{5}{2}}=\sqrt{2,5}[/latex] [latex]4)\sqrt{9,9}-\sqrt{4,4}-\sqrt{1,1}=\sqrt{9\cdot 1,1}-\sqrt{4\cdot 1,1}-\sqrt{1,1}=\\\\=3\sqrt{1,1}-2\sqrt{1,1}-\sqrt{1,1}=0[/latex] [latex]5)\; \; a \geq 0\; ,\; \; a\sqrt{14}=\sqrt{14a^2}\\\\b\ \textless \ 0\; ,\; \; b\sqrt5=-\sqrt{5b^2}\\\\c^2\sqrt2=\sqrt{2c^4}\\\\-d^4\sqrt2=-\sqrt{2d^8}\\\\x \geq 0\; ,\; \; x^3\sqrt{14}=\sqrt{14x^6}\\\\y\ \textless \ 0\; ,\; \; -y^5\sqrt{21}=\sqrt{21y^{10}}[/latex]

9. [latex]a)8 \sqrt{0.1} -2 \sqrt{0.4}-0.5 \sqrt{40} =8 \sqrt{0.1} -2 \sqrt{4*0.1}-0.5 \sqrt{400*0.1} = \\ \\ 8 \sqrt{0.1} -2 \sqrt{2^2}* \sqrt{0.1} -0.5 \sqrt{20^2}* \sqrt{0.1} = \\ \\ 8 \sqrt{0.1}-2*2* \sqrt{0.1}-0.5*20* \sqrt{0.1} =8 \sqrt{0.1}-4 \sqrt{0.1}-10 \sqrt{0.1}= \\ \\ -6 \sqrt{0.1} [/latex] [latex]b) \sqrt{ \frac{2}{3} } + \sqrt{ \frac{50}{3} } - \sqrt{6}= \sqrt{ \frac{2}{3} } + \sqrt{ \frac{25*2}{3} } - \sqrt{6}=\sqrt{ \frac{2}{3} } + 5\sqrt{ \frac{2}{3} } - \sqrt{6}= \\ \\ 6\sqrt{ \frac{2}{3} } - \sqrt{6}=6\sqrt{ \frac{2*3}{3^2} } - \sqrt{6}= \frac{6}{3}* \sqrt{6}- \sqrt{6}= 2 \sqrt{6}- \sqrt{6}= \sqrt{6} [/latex] [latex]B) \sqrt{10}- \sqrt{ \frac{605}{8} }+ \frac{99}{4}* \sqrt{ \frac{10}{121}}= \sqrt{10}- \sqrt{ \frac{605*8}{8^2} }+ \frac{99}{4}* \sqrt{ \frac{10}{11^2}}= \\ \\ \sqrt{10}- \frac{1}{8}* \sqrt{ 4840}+ \frac{99}{4*11}* \sqrt{ 10}= \sqrt{10}- \frac{1}{8}* \sqrt{ 484*10}+ \frac{9}{4}* \sqrt{ 10}= \\ \\ \frac{13}{4} *\sqrt{10}- \frac{1}{8}* \sqrt{ 22^2*10}=\frac{13}{4} *\sqrt{10}-\frac{22}{8}* \sqrt{10}=\frac{13}{4} *\sqrt{10}-\frac{11}{4}* \sqrt{10} \\ \\ = \frac{2}{4}* \sqrt{10}= [/latex] [latex] \frac{ \sqrt{10}}{2} [/latex] [latex]s) \sqrt{9.9}- \sqrt{4.4}- \sqrt{1.1}= \sqrt{9*1.1}- \sqrt{4*1.1}- \sqrt{1.1}= \\ \\ \sqrt{3^2*1.1}- \sqrt{2^2*1.1}- \sqrt{1.1}=3*\sqrt{1.1}- 2*\sqrt{1.1}- \sqrt{1.1}=0[/latex] 10. [latex]a)a \sqrt{14}= \sqrt{a^2} * \sqrt{14} = \sqrt{14a^2} \\ \\ b)b \sqrt{5}=- \sqrt{b^2}* \sqrt{5}= - \sqrt{5b^2} \\ \\ B)c^2 \sqrt{2}= \sqrt{c^4}* \sqrt{2}= \sqrt{2c^4} [/latex] [latex]s)-d^4 \sqrt{2}=- \sqrt{(d^4)^2}* \sqrt{2}=- \sqrt{2d^8} \\ \\ g) x^{3}* \sqrt{14}= \sqrt{x^6}* \sqrt{14}= \sqrt{14x^6} \\ \\ e)- y^5* \sqrt{21}=\sqrt{(y^{5})^2}* \sqrt{21}= \sqrt{21y^{10}} [/latex]