Question

6 Fie a, b, c aparține lui R+, cu a•b = 6, b• c = 12, c•a = 8. Aflați a• b• c şi a²b²c + ab²c² + a²bc² .​

Answer 1

Date:

$a,b,c \in R\\a > 0,b > 0,c > 0$

$ab=6\\bc=12\\ca=8$

Se cer:

$a)\;abc=?\\b)\;a^2b^2c+ab^2c^2+a^2bc^2=?$

Rezolvare:

a)  Inmultim relatiile primite:

$(ab)(bc)(ca)=6*12*8\\(abc)^2=6*6*2*2*4\\(abc)^2=(6*2*2)^2\\(abc)^2=24^2$

$abc=24 \;\;sau\;\;abc=-24$

Tinem cont ca numerele sunt pozitive => Prin inmultirea a 3 numere pozitive nu putem  obtine un numar negativ => abc=24

b) Prelucram cerinta:

$a^2b^2c+ab^2c^2+a^2bc^2=abc(ab+bc+ac)\\a^2b^2c+ab^2c^2+a^2bc^2=24*(6+12+8)\\a^2b^2c+ab^2c^2+a^2bc^2=24*26\\a^2b^2c+ab^2c^2+a^2bc^2=624$

Raspuns:

$abc=24\\a^2b^2c+ab^2c^2+a^2bc^2=624$