bjr
Q1
5x7 + 1 = 35 + 1 = 36
multiple de 4 (4x9=36)
Q2ab
développer (2x+1) (2x+3) + 1 = 4x² + 6x + 2x + 3 + 1 = 4x² + 8x + 3 + 1=
4x² + 8x + 4 = 4 (x² + 2x + 1) => donc multiple de 4
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bjr
Q1
5x7 + 1 = 35 + 1 = 36
multiple de 4 (4x9=36)
Q2ab
développer (2x+1) (2x+3) + 1 = 4x² + 6x + 2x + 3 + 1 = 4x² + 8x + 3 + 1=
4x² + 8x + 4 = 4 (x² + 2x + 1) => donc multiple de 4