N. Belliard
Compléments des cours
2NDE 8
Série d'exercices AP01
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\[A(x) = 3 - (2x + 3) = \cancel 3 - 2x - \cancel 3 = -2x.\] \[B(t) = 5t + (-3t + 5) = 5t -3t +5 = 2t + 5.\] \[C(x) = 5 + (5 - 6x) = 5 + 5 - 6x = 10 - 6x.\] \[D(t) = 2 -(-2t + 9) = 2 + 2t - 9 = 2t - 7.\] \[E(x) = 3x + (-6x + 12) = 3x - 6x + 12 = -3x + 12.\] \[F(t) = -6t -(-9t - 11) = -6t + 9t + 11 = 3t + 11.\]
\[\begin{aligned} A(y) &= 2(y-3) + 5(2y - 5)& \\ &=2y - 2\cdot3 + 5\cdot 2y - 5\cdot 5& \\ &=2y - 6 + 10y - 25& \\ &=12y -31.& \end{aligned}\] \[\begin{aligned} B(t) &= 6(5t + 3) - 8(-3+6t)& \\ &=6\cdot 5t + 6\cdot 3 -8 \cdot (-3) - 8\cdot 6t& \\ &=30t + 18 + 24 - 48t& \\ &=-18t + 42.& \end{aligned}\] \[\begin{aligned} C(x) &= -10(4-3x) + (3x + 1)\cdot 6& \\ &=-10\cdot 4 -10\cdot(-3x) + 3x\cdot 6 + 1\cdot 6& \\ &=-40 +30x + 18x + 6& \\ &=48x - 34.& \end{aligned}\] \[\begin{aligned} D(s) &=-5(-2s + 7)-9(3 - 11s)& \\ &=-5\cdot(-2s) -5\cdot 7 - 9\cdot 3 -9\cdot(-11s)& \\ &=10s - 35 - 18 + 99s& \\ &=109s - 53.& \end{aligned}\]
\[\begin{aligned} A(x) &= (x+2)(5x+3)& \\ &=x\cdot 5x + x\cdot 3 + 2\cdot 5x + 2\cdot 3& \\ &=5x^2 + 3x + 10x + 6& \\ &=5x^2 + 13x + 6.& \end{aligned}\] \[\begin{aligned} B(x) &= (4x+7)(x+2)& \\ &=4x\cdot x + 4x \cdot 2 + 7 \cdot x + 7\cdot 2& \\ &=4x^2 + 8x + 7x + 14& \\ &=4x^2 + 15x + 14.& \end{aligned}\] \[\begin{aligned} C(x) &=(7+3x)(2x+3)& \\ &=7\cdot 2x + 7\cdot 3 + 3x\cdot 2x + 3x\cdot 3& \\ &=14x + 21 + 6x^2 + 9x& \\ &=6x^2 + 23x + 21.& \end{aligned}\] \[\begin{aligned} D(t) &= (10-8t)(9+2t)& \\ &=10\cdot 9 + 10 \cdot 2t - 8t\cdot 9 -8t \cdot 2t& \\ &=90 + 20t - 72t - 16t^2& \\ &=-16t^2 -52t + 90.& \end{aligned}\]
\[\begin{aligned} A(x)&=(x-1)(x-2) + (x-5)& \\ &=x\cdot x + x \cdot(-2) -1\cdot x -1 \cdot (-2) + x - 5& \\ &=x^2 -2x -x + 2 + x - 5& \\ &=x^2 - 2x - 3.& \end{aligned}\] \[\begin{aligned} B(x) &=2x + (x-3)(x+5)& \\ &=2x + x\cdot x + x\cdot 5 -3\cdot x -3\cdot 5& \\ &=2x + x^2 + 5x -3x - 15& \\ &=x^2 + 4x - 15.& \end{aligned}\] \[\begin{aligned} C(x) &= 2x - (x+3)(x-1)& \\ &=2x - \left(x\cdot x + x\cdot (-1) + 3\cdot x + 3\cdot (-1)\right)& \\ &=2x - (x^2 - x + 3x - 3)& \\ &=2x - x^2 + x - 3x + 3& \\ &=-x^2 -x + 3.& \end{aligned}\] \[\begin{aligned} D(t) &=6t^2 - (2t -3)(3t-1)& \\ &=6t^2-\left(2t\cdot3t + 2t\cdot(-1) - 3\cdot 3t -3\cdot(-1)\right)& \\ &=6t^2 - (6t^2 - 2t - 9t + 3)& \\ &=6t^2 - 6t^2 + 2t + 9t - 3& \\ &=11t - 3.& \end{aligned}\] \[\begin{aligned} E(x) &= -2x^2 - (1-2x)(x + 1)& \\ &=-2x^2 -\left(1\cdot x + 1\cdot 1 -2x \cdot x - 2x \cdot 1\right)& \\ &=-2x^2 - (x + 1 -2x^2 - 2x)& \\ &=-2x^2 -x - 1 + 2x^2 - 2x& \\ &=-3x - 1.& \end{aligned}\]
\[A(x) = 12x + 20 = {\color{DarkRed}4}\cdot 3x + {\color{DarkRed}4}\cdot 10 = 4(3x + 4).\] \[B(x) = 18x^2 + 24x = 6x\cdot 3x + 6x \cdot 4 = 6x(3x + 4).\] \[C(x) = 75 - 50t = 25\cdot 3 - 25 \cdot 2t = 25(3 - 2t).\] \[D(t) = 21t - 28t^2 = 7t \cdot 3 - 7t \cdot 4t = 7t(3 - 4t).\] \[E(x) = 24x + 40x^2 = 8x\cdot 3 + 8x \cdot 5x = 8x(3+5x).\] \[F(t) = 18t^2 - 45t = 9t\cdot 2t - 9t\cdot 5 = 9t(2t - 5).\]
1. L'aire colorée peut être obtenue en ôtant de l'aire du grand rectangle l'aire du petit rectangle : \[23(2x+8) - x(x+5).\]
2. Développons et réduisons : \[\begin{aligned} &23(2x+8) - x(x+5)& \\ =&23\cdot 2x + 23\cdot 8 - x \cdot x - x\cdot 5 \\ =&46x + 184 - x^2 - 5x& \\ =&-x^2 + 41x + 184.& \end{aligned}\]
1. \[\begin{aligned} &10 \rightarrow 10+2 = 12 \rightarrow 4\times 12 = 48 \rightarrow 48-8 = 40\;;& \\ &32 \rightarrow 32+2 = 34 \rightarrow 4\times 34 = 136 \rightarrow 136 - 8 = 128\;;& \\ &{\small -9 \rightarrow -9+2 = -7 \rightarrow 4\times (-7) = -28 \rightarrow -28 - 8 = -36.}& \end{aligned}\] Il semble que le résultat final soit le quadruple du nombre de départ.
2. Si $x$ est le nombre de départ, on calcule successivement \[x\rightarrow x + 2 \rightarrow 4(x+2) \rightarrow 4(x+2) - 8.\] Développons et réduisons cette expression: \[4(x+2) - 8 = 4x + 4\cdot 2 - 8 = 4x + 8 - 8 = 4x.\] Donc notre conjecture est vérifiée.
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