(2√5-√3)(√3+3√5) =
2√5*√3+2√5*3√5-√3*√3-√3*3√5 =
2√15+6√25-√9-3√15 =
2√15-3√15+6*5-3 =
-√15+30-3 =
27-√15
![x^2=t; t^2-17t+16=0](https://tex.z-dn.net/?f=x%5E2%3Dt%3B+t%5E2-17t%2B16%3D0)
D=289-64=225
t=(17+15)/2=16 t=(17-15)/2=1
x^2=16 ==> x=+-4
x^2=1 ==>x=+-1
(4-х)-(5-2х), при х=2
(4-2)-(5-4)
2-1=1
Ответ 1
14*a^(2/5)-10*(a^(1/5))^2=
=14*a^(2/5)-10*a^(2/5)=
=4*a^(2/5)