F.2 maths questions x2...thz

1.分解為因式
(a)
3a(4b - 1) + 2a(1 - 4b)
= 3a(4b - 1) - 2a(4b - 1)
= (3a - 2a)(4b - 1)
= a(4b - 1)

(b)
(5x - 2)(y - 3z) - (2 - 5x)(3y - z)
= (5x - 2)(y - 3z) + (5x - 2)(3y - z)
= (5x - 2) [(y - 3z) + (3y - z)]
= (5x - 2)(y - 3z + 3y - z)
= (5x - 2)(4y - 4z)
= 4(5x - 2)(y - z)

(c)
am + bm - an - bn - m + n
= am - an + bm - bn - m + n
= (am - an) + (bm - bn) - (m - n)
= a(m - n) + b(m - n) - (m - n)
= (a + b - 1)(m - n)

(d)
ax + 2y - ay - 2z - 2x + az
= ax - ay + az - 2x + 2y - 2z
= (ax - ay + az) - (2x - 2y + 2z)
= a(x - y + z) - 2(x - y + z)
= (a - 2)(x - y + z)


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2.化簡代數分式
(a)
(4x + 2x²)/(2 + x)²
= 2x(2 + x)/(2 + x)²
= 2x/(2 + x)

(b)
(2a - 2b)/3a² ÷ (b - a)/9ab
= 2(a - b)/3a² x (-9ab)/(a - b)
= -6b/a