a fost răspuns

1. aflati numerele rationale pozitive a, b, c pentru care 6a=8b=12c si 4a-5b+6c=91
2.  determinati pe x din proportia √48/4=x+1/√3  (/= supra)
3. Daca a=(√3 -1)(√5 +2) si b=(√3 +1)(√5 -2), calculati media geometrica a celor 2 numere.

Răspuns :

1)6a=8b=12c=k
6a=k⇒  a=k/6
8b=k⇒  b=k/8
12c=k⇒  c=k/12
4a-5b+6c=91 inlocuim
4k/6-5k/8+6k/12=91 aducem la acelasi numitor care este 24
(16k-15k+12k)/24=91
13k/24=91
k=91·24/13
k=168
a=k/6=168/6
a=28
b=k/8=168/8=21
c=k/12=168/12=14
2)⇒√48·√3=4(x+1)
√144=4x+1
12-1=4x
11=4x
x=11/4=2,75
3)⇒a=√15+2√3-√5-2
b=√15-2√3+√5-2
Mg=√ab
a·b=(√15+2√3-√5-2)(√15-2√3+√5-2)=
     =15-2√45+√75-2√15+2√45-6+2√15-4√3-√75+2√15-5+2√5-2√15+4√3-2√5+4=15-6-5+4=8
Mg=√8=2√2