[20分]六條中二數學[恆等式]及[因式分解]問題

1.
L.H.S. = u(v - w)2 + v(u - w)2
= u(v2 - 2vw + w2) + v(u2 - 2uw + w2)
= uv2 - 2uvw + 2uw2 + u2v - 2uvw + vw2
= uv2 - 4uvw + 2uw2 + u2v + vw2

R.H.S. = w(u - v)2
= w(u2 - 2uv + v2)
= u2w - 2uvw + v2w

Hence, L.H.S. ≠ R.H.S.


2.
(x - A)(x - 2) = x2 + Bx - 6
x2 - 2x - Ax + 2A = x2 + Bx - 6
x2 - (2 + A)x + 2A = x2 + Bx - 6

Compare the constant terms:
2A = -6
A = -3

Compare the x terms:
-(2 + A)x = Bx
-(2 - 3) = B
B = 1

Ans: A = -3 and B = 1
(The given answer of B is incorrect.)


3.
i.
x(y + 1)2 - 4xy - 4x
= x(y + 1)2 - 4x(y + 1)
= x(y + 1)[(y + 1) - 4]
= x(y + 1)(y - 3)

ii.
ky(2k - h)(2x - 2y) - y(2k - h)2(x - y) + 2h(2h - 4k)(x - y)2
= 2ky(2k - h)( x - y) - y(2k - h)2(x - y) + 4h(h - 2k)(x - y)2
= 2ky(2k - h)( x - y) - y(2k - h)2(x - y) - 4h(2k - h)(x - y)2
= (2k - h)(x - y)[2ky - y(2k - h) - 4h(x - y)]
= (2k - h)(x - y)[2ky - 2ky + hy - 4hx + 4hy]
= (2k - h)(x - y)(5hy - 4hx)
= h(2k - h)(x - y)(5y - 4x)


4.
i.
x3 - x2y + xy2
= x(x2 - xy + y2)

ii.
x2y - xy2 + y3
= y(x2 - xy + y2)

iii.
x3 + y3
= (x3 - x2y + xy2) + (x2y - xy2 + y3)
= x(x2 - xy + y2) + y(x2 - xy + y2)
= (x + y)(x2 - xy + y2)


5.
i.
-m2(r - 1) + m - m2
= -m2r + m2 + m - m2
= -m2r + m
= m - m2r
= m(1 - mr)

ii.
m(r - 1)2 - (r - 1) - m(1 - r)
= m(r - 1)2 - (r - 1) + m(r - 1)
= (r - 1)[m(r - 1) - 1 + m]
= (r - 1)(mr - m - 1 + m)
= (r - 1)(mr - 1)

iii.
[-m2(r - 1) + m - m2] + [m(r - 1)2 - (r - 1) - m(1 - r)]
= m(1 - mr) + (r - 1)(mr - 1)
= m(1 - mr) - (r - 1)(1 - mr)
= (1 - mr)[m - (r - 1)]
= (1 - mr)(m - r + 1)