數學就考試 問題

1.
3x³ - 2x² + 12x ≡ x(Ax - 1)(x - B) + Cx(x + B)
3x³ - 2x² + 12x ≡ x(Ax² - ABx - x + B) + Cx² + BCx
3x³ - 2x² + 12x ≡ Ax³ - ABx² - x² + Bx + Cx² + BCx
3x³ - 2x² + 12x ≡ Ax³ + (C- AB - 1)x² + (B + BC)x

Compare the x³ terms on the both sides :
A = 3

Compare the x² terms on the both sides :
C - AB - 1 = -2
C - 3B = -1 ...... [1]

Compare the x terms on the both sides :
B + BC = 12
BC = 12 - B
C = (12 - B)/B ...... [2]

Put [2] into [1] :
[(12 - B)/B] - 3B = -1
(12 - B - 3B²)/B = -1
12 - B - 3B² = -B
B² = 4
B = 2 or B = -2

Hence, A = 3, B = 2, C = 5
or A =3, B = -2, C = -7


2.
2x³ - x² + 5x + 6 ≡ (x + 2)(Ax² + Bx + C) + D
2x³ - x² + 5x + 6 ≡ Ax³ + Bx² + Cx + 2Ax² + 2Bx + 2C + D
2x³ - x² + 5x + 6 ≡ Ax³ + (2A + B)x² + (2B + C)x + (2C + D)

Compare the x³ terms on the both sides :
A = 2

Compare the x² terms on the both sides :
2A + B = -1
2(2) + B = -1
B = -5

Compare the x terms on the both sides :
2B + C = 5
2(-5) + C = 5
C = 15

Compare the constant terms on the both sides :
2C + D = 6
2(15) + D = 6
D = -24


3.
(a - b)³ - a³ + ab²
= [(a - b)³ - a³] + ab²
= [(a - b) - a] [(a - b)² +(a - b)a + a²] + ab²
= [a - b - a] [a² - 2ab + b² + a² - ab + a²] + ab²
= -b(3a² - 3ab + b²) + ab²
= -b[(3a² - 3ab + b²) - ab]
= -b[3a² - 4ab + b²]
= -b(3a - b)(a - b)

Alternative method :
(a - b)³ - a³ + ab²
= a³ - 3a²b + 3ab² - b³ - a³ + ab²
= -3a²b + 4ab² - b³
= -b(3a² - 4ab + b²)
= -b(3a - b)(a - b)


4.
[(3√a)/2] - [a/√(4a)]
= [(3/√a)/2] - [(√a)²/2√a]
= [(3/√a)/2] - [(√a)/2]
= [(3√a) - (√a)]/2
= 2(√a)/2
= √a