✔ 最佳答案
1.
a 及 b 為二次方程3x² - 3x - 11 = 0 的根:
兩根之和: a + b = -(-3/3) = 1
兩根之積: ab =-11/3
(a)
(1/a) + (1/b)
= (a + b) / ab
= 1 / (-11/3)
= -3/11
(b)
a² + b²
= (a + b)² - 2ab
= 1 - 2(-11/3)
= 1 + 22/3
= 25/3
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2.
a 及 b 為二次方程 5x² + 3x - 2 = 0 的根:
兩根之和: a + b = -3/5
兩根之積: ab = -2/5
(a - b)²
= (a + b)² - 4ab
= (-3/5)² - 4(-2/5)
= (9/25) + 40/25
= 49/25
故此 a - b
= ±7/5
(a)
a³ - b³
= (a - b)(a² + ab + b²)
= (a - b) [(a + b)² - ab]
= (±7/5) [(-3/5)² - (-2/5)]
= (±7/5) [(9/25) + (10/25)]
= ±133/125
(b)
(a/b²) - (b/a²)
= (a³ - b³) / (ab)²
= (±133/125) / (-2/5)²
= (±133/125) (25/4)
= ±133/20
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3.
y = (x + 1)² + k
y = x² + 2x + 2
y = x² + 2x + 3
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4.
線段 AB 的方程式:
y/(x - 7) = -5/(2 - 7)
x - y - 7 = 0
求 2x + y -5 = 0 與 x - y - 7 = 0 的交點:
2x + y - 5 = 0 ... [1]
x - y - 7 = 0 ... [2]
[1] + [2] :
3x - 12 = 0
x = 4
兩線相交點的 x 坐標 = 4
把 2x + y - 5 = 0 把線段 AB 所分割的比
= (4 - 2) : (7 - 4)
= 2 : 3
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5.
直線一:斜率為 1,y 截距為 1
y = x + 1
x - y + 1 = 0
直線二:斜率為 -1,y 截距為 1
y = -x + 1
x + y - 1 = 0
6.
(a)
f(x) = x¹⁵ + 16¹⁵
f(-16)
= (-16)¹⁵ + 16¹⁵
= -16¹⁵ + 16¹⁵
= 0
f(-16) = 0,根據因式定理:
f(x) 可被 (x + 16) 整除。
即 (x + 16) 為 f(x) 的因式。
(b)
由 (a) 得證︰ (x¹⁵ + 16¹⁵) 可被 (x + 16) 整除。
令 x = 4:
(4¹⁵ + 16¹⁵) 可被 (4 + 16) 整除。
即 [(2²)¹⁵ + (4²)¹⁵] 可被 20 整除。
即 (2³⁰ + 4³⁰) 可被 20 整除。
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7.
sin27° = sin(180° - 27°) = sin153°
sin27° = sin(360° + 27°) = sin387°
sin27° = cos(90° - 27°) = cos63°
sin27° = cos63° = cos(360°- 63°) = cos297°
sin27° = cos63° = cos(360° +63°) = cos423°
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8.
多項式一:
x(x - 1)(x - 2)
= x³ - 3x² + 2x
多項式二:
(x + 1)(x - 1)(x - 2)
= (x² - 1)(x - 2)
= x³ - 2x² - x + 2
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9.
線段 AB 的方程式:
y/(x - 7) = -5/(2 - 7)
x - y - 7 = 0
求 2x + y -5 = 0 與 x - y - 7 = 0 的交點:
2x + y - 5 = 0 ... [1]
x - y - 7 = 0 ... [2]
[1] + [2] :
3x - 12 = 0
x = 4
兩線相交點的 x 坐標 = 4
把 2x + y - 5 = 0 把線段 AB 所分割的比
= (4 - 2) : (7 - 4)
= 2 : 3