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標題:
數學一問用有向數+代數......10點
發問:
唔該幫我列條方程出黎,要計埋,要有步驟!!!!!!!!!!! 1. 2y-[5y+(6-y)]=12 2. 3(2x-1)+4[x+7(x+3)]=30x-34 3. 19a-2[a-6(3-a)]=14 4. 8x-[(3x-4)+2(x+7)]-2(4x+15) 5. 4(2n-3)+3[3n-8(n-2)]=15-4n 6. 2(11x-16)-5[7x-3(4-x)]=7x+2
最佳解答:
1) 2y - [5y + (6 - y)] = 12 2y - [5y - y + 6] = 12 2y - 4y - 6 = 12 - 2y = 18 ∴ y = - 9 2) 3(2x - 1) + 4[x + 7(x + 3)] = 30x - 34 6x - 3 + 4[x + 7x + 21] = 30x - 34 6x - 3 + 32x + 84 = 30x - 34 8x = - 115 ∴ x = - 115/8 3) 19a - 2[a - 6(3 - a)] = 14 19a - 2[a - 18 + 6a] = 14 19a - 14a + 36 = 14 5a = - 22 ∴ a = - 22/5 4) 8x - [(3x - 4) + 2(x + 7)] = 2(4x + 15) 8x - [3x - 4 + 2x + 14] = 8x + 30 - 5x - 10 = 30 - 5x = 40 ∴ x = - 8 5) 4(2n - 3) + 3[3n - 8(n - 2)] = 15 - 4n 8n - 12 + 9n - 24n + 48 = 15 - 4n - 7n + 36 = 15 - 4n 21 = 3n 3n = 21 ∴ n = 7 6) 2(11x - 16) - 5[7x - 3(4 - x)] = 7x + 2 22x – 32 – 35x + 60 – 15x = 7x + 2 - 28x +28 = 7x +2 26 = 35x 35x = 26 ∴ x = 26/35 希望幫到您!!!
數學一問用有向數+代數......10點
發問:
唔該幫我列條方程出黎,要計埋,要有步驟!!!!!!!!!!! 1. 2y-[5y+(6-y)]=12 2. 3(2x-1)+4[x+7(x+3)]=30x-34 3. 19a-2[a-6(3-a)]=14 4. 8x-[(3x-4)+2(x+7)]-2(4x+15) 5. 4(2n-3)+3[3n-8(n-2)]=15-4n 6. 2(11x-16)-5[7x-3(4-x)]=7x+2
最佳解答:
1) 2y - [5y + (6 - y)] = 12 2y - [5y - y + 6] = 12 2y - 4y - 6 = 12 - 2y = 18 ∴ y = - 9 2) 3(2x - 1) + 4[x + 7(x + 3)] = 30x - 34 6x - 3 + 4[x + 7x + 21] = 30x - 34 6x - 3 + 32x + 84 = 30x - 34 8x = - 115 ∴ x = - 115/8 3) 19a - 2[a - 6(3 - a)] = 14 19a - 2[a - 18 + 6a] = 14 19a - 14a + 36 = 14 5a = - 22 ∴ a = - 22/5 4) 8x - [(3x - 4) + 2(x + 7)] = 2(4x + 15) 8x - [3x - 4 + 2x + 14] = 8x + 30 - 5x - 10 = 30 - 5x = 40 ∴ x = - 8 5) 4(2n - 3) + 3[3n - 8(n - 2)] = 15 - 4n 8n - 12 + 9n - 24n + 48 = 15 - 4n - 7n + 36 = 15 - 4n 21 = 3n 3n = 21 ∴ n = 7 6) 2(11x - 16) - 5[7x - 3(4 - x)] = 7x + 2 22x – 32 – 35x + 60 – 15x = 7x + 2 - 28x +28 = 7x +2 26 = 35x 35x = 26 ∴ x = 26/35 希望幫到您!!!
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