Question

Verificati relatia: media armonica (doar cele incercuite)
Multumesc!

Answer 1

c)  Ma=(5+7,2):2=12,2:2=6,1
Mg=√5·7,2=√36=6
Mh=(2·5·7,2) /12,2=72/12,2=36/6,1
Mh<Mg<Ma

d) 28√3 si 12√3
Ma=(28√3+12√3):2=40√3:2=20√3
Mg=√2²·2²·3²·21=12√21
Mh=(2·28√3·12√3)28√3+12√3=252/5√3

e)√2 si  √8=2√2
Ma=(√2+2√2):2=3√2/2
Mg=√√16=√4=2
Mh=4√2/3√2=4/3=1 si 1/3

Answer 2

c)ma=( 5+7,2)/2=12,2/2=6,1
mg=√5×7,2=√36=6
mh=(2×5×7,2)/5+7,2=72:12,2=5,9
5,9<=mg<=6,1
d)ma=40√3/2=20√3
mg=√28×12×3=√4×7×4×3×3=12√7
mh=672/40√3=336√3/20=84√3/5
29,04<=31,68<=34,6
e)ma=(√2+2√2)/2=3√2/2
mg=√16=4
mh=2√2×√8/3√2=8√2/3
3,76<=4<=2,11 fais
f)1 intreg 1/7=8/7
3/125
ma=(8/7+3/125):2=(125×8+3×7)/250=1021/250=4,082
mg=√8/7×3/125=√4×6/35×25=2√6/5√35=0,.....
mh=(48/875):1021/125=48/875×125/1021=48/(7×1021)=0,0006
mh<=mg<=ma