Réponse:
Bonjour,
tu dois utiliser k×a+k×b=k×(a+b)
[tex]a = 3 {x}^{2} + 3 = 3 \times {x}^{2} + 3 \times 1 = 3 \times ( {x}^{2} + 1)[/tex]
[tex]b = 7 - 7x = 7 \times 1 + 7 \times ( - 1) \times x \\ = 7 \times 1 + 7 \times ( - x) = 7 \times (1 + ( - x)) \\ = 7 \times (1 - x)[/tex]
[tex]c = 21t - 7 = 7 \times 3 \times t + 7 \times ( - 1) \\ = 7 \times (3 \times t + ( - 1)) \\ = 7 \times (3t - 1)[/tex]
[tex]d = 10 \times t + 1 \times t = t \times 10 + t \times 1 \\ = t \times (10 - 1) = t \times 9 = 9t[/tex]
[tex]e = 3 {x}^{2} + x = 3 \times x \times x + 1 \times x \\ = x \times 3x + x \times 1 = x \times (3x + 1)[/tex]
[tex]f = {n}^{2} - n = n \times n + n \times ( - 1) \\ = n \times (n + ( - 1)) = n \times (n - 1)[/tex]
