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Qual o valor de x na equação: (x+3)! + (x+2)! = 8(x+1)!
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| f(x: Inteiro): Inteiro = escolha x | |
| caso 0 => 1 | |
| caso n => n * f(n - 1) | |
| fim | |
| solução = | |
| para x de 0 até 10 se f(x + 3) + f(x + 2) == 8 * f(x + 1) gere x fim | |
| escreva solução |
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Resolvendo algebricamente:
(x+3)! + (x+2)! = 8 * (x+1)!
(x+3)(x+2)(x+1)! + (x+2)(x+1)! = 8 * (x+1)!
(x+3)(x+2)+(x+2) = 8 , x>=0
(x+4)_(x+2) = 8
x^2+6_x = 0
x * (x + 6) = 0
x' = 0
x'' = -6 (viola a condição x>=0)