Question

Exercitiul 14- e si f
Exercitiul 15-tot

Answer 1

14e)
m=√3+2√2
n=√3-2√2
ma=(√3+2√2+√3-2√2)/2=2√3/2=√3

14f)
m=1/(√3+√2)
n=1/(√3-√2)
ma=[1/(√3+√2)+1/(√3-√2)]/2=(√3-√2+√3+√2)/(√3+√2)(√3-√2)]/2=
=[2√3)/(3-2)]/2=
=(2√3)/2=√3


15a)
a=√3+√2
b=√3-√2
mg=√(√3+√2)(√3-√2)=√(3-2)=√1=1

b)
a=3√5+4√3
b=3√5-4√3
mg=√(3√5+4√3)(3√5-4√3)=√(9*5-4*3)=√(45-48)=√(-3)     

c)
a=3√2+2√5 =2√5+3√2
b=2√5-3√2
mg=√(2√5+3√2)(2√5-3√2)=√(4*5-9*2)=√(20-18)=√2

d)
a=√15+√16
b=-√16+√15 =√15-√16
mg=√(√15+√16)(√15-√16 )=√(15-16)=√(-1)

e)
a=√(3+2√2)
b=√(3-2√2)
mg=√(√(3+2√2)√(3-2√2)=√√(9-4*2)=1

f)
a=1/(√3+√2)
b=1/(√3-√2)
mg=√(1/(√3+√2)(√3-√2)=√1/(3-2)=√1/1=1