Question

Пожалуйста помогите решить задания . Желательно 6 заданий. Даю макс. кол-во баллов.

Answer 1

$1)\\a)\;\sqrt{169}-3\sqrt{0,36}=13-3\cdot0,6=13-1,8=11,2\\b)\;\sqrt{3\frac17}\cdot\sqrt{\frac{7}{88}}=\sqrt{\frac{22}7}\cdot\sqrt{\frac7{88}}=\sqrt{\frac{22}7\cdot\frac7{88}}=\sqrt{\frac14}=\frac12\\\\c)\;\sqrt{2^6\cdot5^4}=\sqrt{(2^3\cdot5^2)^2}=2^3\cdot5^2=8\cdot25=200\\d)\;\frac{\sqrt{500}}{\sqrt{10}\cdot\sqrt{32}}=\frac{\sqrt{500}}{\sqrt{10\cdot32}}=\sqrt{\frac{500}{320}}=\sqrt{\frac{25}{16}}=\frac54=1\frac14$

$2)\\a)\;x^2=13\\x=\pm\sqrt{13}\\\\b)\;x^2+1=0\\x^2=-1\;-\;KOPHEu\;HET.\\\\c)\;\sqrt x=4\\x=\pm2\\\\d)\;\sqrt x=-9\\x=81$

$3)\\a)\;2\sqrt3-\sqrt{48}+\sqrt{75}=2\sqrt3-\sqrt{16\cdot3}+\sqrt{25\cdot3}=2\sqrt3-4\sqrt3+5\sqrt3=3\sqrt3\\\\b)\;\left(\sqrt{63}-\sqrt{28}\right)\cdot\sqrt7=\left(\sqrt{9\cdot7}-\sqrt{4\cdot7}\right)\cdot\sqrt7=\left(3\sqrt7-2\sqrt7\right)\cdot\sqrt7=\sqrt7\cdot\sqrt7=7\\\\c)\;(3\sqrt6-4)^2=(3\sqrt6)^2-2\cdot4\cdot3\sqrt6+4^2=9\cdot6-24\sqrt6+16=70-24\sqrt6\\\\d)\;(2\sqrt7-3\sqrt2)(2\sqrt7+3\sqrt2)=(2\sqrt7)^2-(3\sqrt2)^2=4\cdot7-9\cdot2=28-18=10$

$4)\\a)\;3\sqrt7=\sqrt{9\cdot7}=\sqrt{63}\\7\sqrt3=\sqrt{49\cdot3}=\sqrt{147}\\\sqrt{63}<\sqrt{147}\Rightarrow3\sqrt7<7\sqrt3\\\\b)\;6\sqrt{\frac7{18}}=\sqrt{36\cdot\frac7{18}}=\sqrt{2\cdot7}=\sqrt{14}\\\frac13\sqrt{108}=\sqrt{\frac19\cdot108}=\sqrt{36}\\\sqrt{14}<\sqrt{36}\Rightarrow6\sqrt{\frac7{18}}<\frac13\sqrt{108}$

$\ldots\\6)\\a)\;\frac{49-b}{7+\sqrt b}=\frac{(7-\sqrt b)(7\sqrt b)}{7+\sqrt b}=7-\sqrt b\\\\b)\;\frac{\sqrt b}{b+2\sqrt b}=\frac{\sqrt b}{\sqrt b(\sqrt b+2)}=\frac1{\sqrt b+2}\\\\c)\;\frac{9-b}{9-6\sqrt b+b}=\frac{(3-\sqrt b)(3+\sqrt b)}{(3-\sqrt b)^2}=\frac{3+\sqrt b}{3-\sqrt b}$

$7)\\a)\;\frac4{3\sqrt5}=\frac{4\cdot\sqrt5}{3\sqrt5\cdot\sqrt5}=\frac{4\sqrt5}{3\cdot5}=\frac{4\sqrt5}{15}\\\\b)\;\frac{12}{\sqrt{15}+3}=\frac{12\cdot(\sqrt{15}-3)}{(\sqrt{15}+3)(\sqrt{15}-3)}=\frac{12(\sqrt15-3)}{15-9}=\frac{12(\sqrt{15}-3)}6=2\sqrt{15}-6$

$8)\\a)\;\sqrt{13a^2}=-a\sqrt{13}\\b)\;\sqrt{63a^4}=\sqrt{7\cdot9a^4}=3a^2\sqrt7\\c)\;\sqrt{-a^3}=-a\sqrt a\\d)\;\sqrt{-a^3c^6}=ac^3\sqrt a$

$9)\\\sqrt{\left(8-\sqrt{47}\right)^2}+\sqrt{\left(6-\sqrt{47}\right)^2}\\\\8-\sqrt{47}=\sqrt{64}-\sqrt{47}>0\\6-\sqrt{47}=\sqrt{36}-\sqrt{47}<0\\\\\sqrt{\left(8-\sqrt{47}\right)^2}+\sqrt{\left(6-\sqrt{47}\right)^2}=8-\sqrt{47}-6+\sqrt{47}=2$