Question

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

Answer 1

$1)\a);sqrt{169}-3sqrt{0,36}=13-3cdot0,6=13-1,8=11,2\b);sqrt{3frac17}cdotsqrt{frac{7}{88}}=sqrt{frac{22}7}cdotsqrt{frac7{88}}=sqrt{frac{22}7cdotfrac7{88}}=sqrt{frac14}=frac12\\c);sqrt{2^6cdot5^4}=sqrt{(2^3cdot5^2)^2}=2^3cdot5^2=8cdot25=200\d);frac{sqrt{500}}{sqrt{10}cdotsqrt{32}}=frac{sqrt{500}}{sqrt{10cdot32}}=sqrt{frac{500}{320}}=sqrt{frac{25}{16}}=frac54=1frac14$

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

$3)\a);2sqrt3-sqrt{48}+sqrt{75}=2sqrt3-sqrt{16cdot3}+sqrt{25cdot3}=2sqrt3-4sqrt3+5sqrt3=3sqrt3\\b);left(sqrt{63}-sqrt{28}right)cdotsqrt7=left(sqrt{9cdot7}-sqrt{4cdot7}right)cdotsqrt7=left(3sqrt7-2sqrt7right)cdotsqrt7=sqrt7cdotsqrt7=7\\c);(3sqrt6-4)^2=(3sqrt6)^2-2cdot4cdot3sqrt6+4^2=9cdot6-24sqrt6+16=70-24sqrt6\\d);(2sqrt7-3sqrt2)(2sqrt7+3sqrt2)=(2sqrt7)^2-(3sqrt2)^2=4cdot7-9cdot2=28-18=10$

$4)\a);3sqrt7=sqrt{9cdot7}=sqrt{63}\7sqrt3=sqrt{49cdot3}=sqrt{147}\sqrt{63}<sqrt{147}Rightarrow3sqrt7<7sqrt3\\b);6sqrt{frac7{18}}=sqrt{36cdotfrac7{18}}=sqrt{2cdot7}=sqrt{14}\frac13sqrt{108}=sqrt{frac19cdot108}=sqrt{36}\sqrt{14}<sqrt{36}Rightarrow6sqrt{frac7{18}}<frac13sqrt{108}$

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

$7)\a);frac4{3sqrt5}=frac{4cdotsqrt5}{3sqrt5cdotsqrt5}=frac{4sqrt5}{3cdot5}=frac{4sqrt5}{15}\\b);frac{12}{sqrt{15}+3}=frac{12cdot(sqrt{15}-3)}{(sqrt{15}+3)(sqrt{15}-3)}=frac{12(sqrt15-3)}{15-9}=frac{12(sqrt{15}-3)}6=2sqrt{15}-6$

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

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