Question

Ex.4 va rog dau coroana! !

Answer 1

√(A±√B)=√(A+C)/2±√(A-C)/2, unde C^2=A^2-B (Formula radicalilor compuși)
√(14-6√5)=√14-√180, C^2=14^2-180=196-180=16, C=√16=4
=√(14+4)/2-√(14-4)/2=
=√18/2-√10/2=
=√9-√5=
=3-√5
a=3-√5-(3+√5)
a=3-√5-3-√5
a=-2√5
a) (a^2-21)^21=((-2√5)^2-21)^21=(20-21)^21=(-1)^2+=-1
b) [-1]=-2