Question

b 6 a² + b² +c² să fie supraunitară​

Answer 1

Răspuns:

$6 a ^ { 2 } + b ^ { 2 } + c ^ { 2 }$

$12a$

Answer 2

Explicație pas cu pas:

a, b, c numere naturale

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

${0}^{2} + {0}^{2} + {1}^{2} < 6$

(a,b,c): (0,0,1), (0,1,0), (1,0,0)

${0}^{2} + {0}^{2} + {2}^{2} < 6$

(a,b,c): (0,0,2), (0,2,0), (2,0,0)

${0}^{2} + {1}^{2} + {1}^{2} < 6$

(a,b,c): (0,1,1), (1,0,1), (1,1,0)

${0}^{2} + {1}^{2} + {2}^{2} < 6$

(a,b,c): (0,1,2), (0,2,1), (1,0,2), (1,2,0), (2,0,1), (2,1,0)

${1}^{2} + {1}^{2} + {1}^{2} < 6$

(a,b,c): (1,1,1)