Question

b) determinati numerele naturale a, b si c astfel incat fractia 6/a²+b²+c² sa fie supraunitara

c) ab+bc+ca/44 sa fie subunitara

d) 2^a+b/3^c sa fie exhiunitara

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Answer 1

Explicație pas cu pas:

b)

$\frac{6}{ {a}^{2} + {b}^{2} + {c}^{2}} > 1 \\ \\ {a}^{2} + {b}^{2} + {c}^{2} < 6$

${0}^{2} + {0}^{2} + {1}^{2} = 0 + 0 + 1 = 1 < 6 \\ {0}^{2} + {1}^{2} + {1}^{2} = 0 + 1 + 1 = 2 < 6 \\ {1}^{2} + {1}^{2} + {1}^{2} = 1 + 1 + 1 = 3 < 6 \\ {0}^{2} + {1}^{2} + {2}^{2} = 0 + 1 + 4 = 5 < 6$

(a b,c) ∈ {(0,0,1); (0,1,0); (1,0,0); (0,1,1); (1,0,1); (1,1,0); (1,1,1,); (0,1,2); (0,2,1); (1,0,2); (1,2,0); (2,0,1); (2,1,0)}

c) a≠0, b≠0, c≠0

$\frac{\overline {ab} + \overline {bc} + \overline {ca}}{44} <1 \\ \\ \overline {ab} + \overline {bc} + \overline {ca} < 44$

$10a + b + 10b + c + 10c + a < 44$

$11a + 11b + 11c < 44$

$11(a + b + c) < 44$

$a + b + c < 4$

=> 1 + 1 + 1 = 3 < 4

=> a = b = c = 1

(a,b,c) = (1,1,1)

d)

$\frac{ {2}^{a + b} }{ {3}^{c} } = 1 \iff {2}^{a + b} = {3}^{c}$

(2,3) = 1

=> a + b = c = 0

=> a = b = c = 0

(a,b,c) = (0,0,0)