Question

Fie abc trei numere rationale pozitive direct proportionale cu 2,3,4 ​

Answer 1

Răspuns:

$a)(a.b.c) \: d.p \: (2.3.4) \\ \frac{a}{2} = \frac{b}{3} = \frac{c}{4} = k \\ \frac{a}{2} = k \: \: \frac{b}{3} = k \: \: \frac{c}{4} = k \\ n = ( \frac{a}{2} + \frac{b}{3} + \frac{c}{4} )( \frac{2}{a} + \frac{3}{b} + \frac{4}{c}) \\ n = (k + k + k)( \frac{1}{k} + \frac{1}{k} + \frac{1}{k} ) \\ n = 3k \times \frac{3}{k} = \frac{9k}{k} = 9apartine \: multimii \: numerelor \: naturale$

$b) \frac{a}{2 } = k \: \: a = 2k \\ \frac{b}{3} = k \: \: b = 3k \\ \frac{c}{4} = k \: c = 4k \\ b = \frac{a + c}{2} \: \: 3k = \frac{2k + 4k}{2} \: \: \: 3k = \frac{6k}{2} \: \: 3k = 3k \: b \: medie \: aritmetica \: a \: numerelor \: a \: si \: c$

Sper sa fie bn nu sunt sigur la b)