Question

Пожалуйста пожалуйста помогите с решением

Answer 1

$1)quad (frac{asqrt2}{(1+a^2)^{-1}}-frac{2sqrt2}{a^{-1}})cdot frac{a^{-3}}{1-a^{-2}}=(asqrt2(1+a^2)-2sqrt2a)cdot frac{1}{a^3(1-frac{1}{a^2})}=\\=asqrt2(1+a^2-2)cdot frac{a^2}{a^3(a^2-1)}= frac{asqrt2(a^2-1)}{a(a^2-1)}=sqrt2\\2)quadsqrt{4+2sqrt3}=sqrt{(1+sqrt3)^2}=1+sqrt3\\sqrt{9-4sqrt5}=sqrt{(sqrt5-2)^2}=sqrt5-2\\sqrt{10+2sqrt{21}}=sqrt{(sqrt3+sqrt7)^2}=sqrt3+sqrt7\\sqrt{a+2sqrt{a-1}}=sqrt{(1+sqrt{a-1})^2}=1+sqrt{a-1}$

$3)quad sqrt{7+4sqrt3}+sqrt{7-4sqrt3}=sqrt{(2+sqrt3)^2}+sqrt{(2-sqrt3)^2}=\\=|2+sqrt3|+|2-sqrt3|=2+sqrt3+2-sqrt3=4\\\sqrt{14+6sqrt5}+sqrt{14-6sqrt5}=sqrt{(3+sqrt5)^2}+sqrt{(3-sqrt5)^2}=\\=|3+sqrt5|+|3-sqrt5|=3+sqrt5+3-sqrt5=6\\\left (sqrt{3+sqrt5}+sqrt{3-sqrt5}right )^4=\\=left ((3+sqrt5)+2sqrt{(3+sqrt5)(3-sqrt5)}+(3-sqrt5)right )^2=\\=left (6+2sqrt{9-5}right )^2=(6+2cdot sqrt4)^2=(6+2cdot 2)^2=10^2=100$