(f.2因式分解)利用平方差或说完全平方的恒等式,展开下列各数式1.(5x-40y)(x/7+8y/7)2.16(-2x/3-y/3)(y/4-x/2)3.[4(3y/2+1/6)]^24.[(a+b)(a-b)]^25.(x+2y-3x)(x+2y+3z)

Question

(f.2 因式分解)
利用平方差或说完全平方的恒等式,展开下列各数式
1.(5x-40y)(x/7+8y/7)
2.16(-2x/3-y/3)(y/4-x/2)
3.[4(3y/2+1/6)]^2
4.[(a+b)(a-b)]^2
5.(x+2y-3x)(x+2y+3z)

Answer 1

1.
(5x-40y)(x/7+8y/7)
=(5/7)(x-8y)(x+8y)
=(5/7)(x^2-64y^2)
=5x^2/7-320y^2/7
2.
16(-2x/3-y/3)(y/4-x/2)
=16(-1/3)(1/4)(2x+y)(y-2x)
=(-4/3)(y^2-4x^2)
=16x^2/3-4y^2/3
3.
[4(3y/2+1/6)]^2
=[4(1/6)(9y+1)]^2
=[(2/3)(9y+1)]^2
=(4/9)(81y^2+18y+1)
=36y^2+8y+4/9
4.
[(a+b)(a-b)]^2
=(a^2-b^2)^2
=a^4-2a^2b^2+b^4
5.
(x+2y-3z)(x+2y+3z)
=[(x+2y)-3z][(x+2y)+3z]
=(x+2y)^2-9z^2
=x^2+4xy+4y^2-9z^2