Question

27x^2-(1-radical3x)^2;x+radical3)^2-(2x+2radical3)^2;(2x+radical5)^2-45

Answer 1

27x²-(1-√3x)²=
=(3√3x+1-√3x)(3√3x-1+√3x)=
=(2√3x+1)(4√3x-1)

(x+√3)²-(2x+2√3)²=
=(x+√3-2x-2√3)(x+√3+2x+2√3)=
=(-x-√3)(3x+3√3)=
=-3(x+√3)(x+√3)=
=-3(x+√3)²

(2x+√5)²-45 =
=(2x+√5)²-(3√5)² =
=(2x+√5-3√5)(2x+√5+3√5) =
=(2x-2√5)(2x+4√5) =
=2×2×(x-√5)(x+2√5)