中二數學4題,20分,急,thz(下)███████████

因式分解下列各式。
1. x3-abx-x2y+ax2-bx2+bxy
=x(x2-ab-xy+ax-bx+by)
=x(x2-xy-bx+by-ab+ax)
=x[x(x-y)-b(x+y)+a(x-b)]
=x[(x+y)(x-b)+a(x-b)]
=x(x+y+a)(x-b)

2. -36p2+24pq-q2-1
?有冇打錯題目?!



利用平方差的恆等式,展開下列各數式。
1. (6a-15)(4a+10)
=[3(2a-5)][2(2a+5)]
=6(2a-5)(2a+5)
=6[(2a)2-52]
=6(4a2-25)
=24a2-150
2. (3r/4+3s/2)(r-2s)
=3/4(r+2s)(r-2s)
=3/4[r2-(2s)2]
=3/4(r2-4s2)
=3/4r2-3s2

1. x3-abx-x2y+ax2-bx2+bxy

=x(x2-ab-xy+ax-bx+by)

=x(x2-xy-bx+by-ab+ax)

=x[x(x-y)-b(x+y)+a(x-b)]

=x[(x+y)(x-b)+a(x-b)]

=x(x+y+a)(x-b)


2. -36p2+24pq-q2-1


1. (6a-15)(4a+10)

=[3(2a-5)][2(2a+5)]

=6(2a-5)(2a+5)

=6[(2a)2-52]

=6(4a2-25)

=24a2-150

2. (3r/4+3s/2)(r-2s)

=3/4(r+2s)(r-2s)

=3/4[r2-(2s)2]

=3/4(r2-4s2)

=3/4r2-3s2

同埋呢3條恆等式:
x^2-y^2≡(x+y)(x-y)
(x+y)^2≡x^2+2xy+y^2
(x-y)^2≡x^2-2xy+y^2

呢3條恆等式I dont know


因式分解下列各式。
1. x^3-abx-x^2y+ax^2-bx^2+bxy
2. -36p^2+24pq-q^2-1

因式分解i dont know


利用平方差的恆等式,展開下列各數式。
1. (6a-15)(4a+10)
2. (3r/4+3s/2)(r-2s)

平方差的恆等式i dont know



because i am form3,所以我唔會識f.2既野