Question

分解因式（x-2)(x-3)+4

要写过程
呵呵，应该是（x+3)
麻烦各位咯

Answer 1

(x-2)(x+3)+4
=(x-2)[(x-2)+5]+4
=(x-2)^2+5(x-2)+4
=[(x-2)+4][(x-2)+1]
=(x+2)(x-1)

提示：本题的关键在于把x-2看做整体。

或者先展开，再因式分解。
(x-2)(x+3)+4
=x^2+x-6+4
=x^2+x-2
=(x-2)(x+1)

Answer 2

⑴7x^2y^2-63x^2z^2
=7x^2(y^2-9z^2)
=7x^2(y+3z)(y-3z)

⑵9ax^2-24axy+16ay^2
=a(9x^2-24xy+16y^2)
=a[(3x)^2-2*3x*4y+(4y)^2]
=a(3x-4y)^2

⑶49(x-2)^2-25(x-3)^2
=[7(x-2)+5(x-3)][7(x-2)-5(x-3)]
=(7x-14+5x-15)(7x-14-5x+15)
=(12x-29)(2x+1)

⑷3a(x^2+4)^2-48ax^2
=3a[(x^2+4)^2-16x^2]
=3a(x^2+4+4x)(x^2+4-4x)
=3a(x^2+2*2x+2^2)(x^2-2*2x+2^2)
=3a(x+2)^2(x-2)^2

追问

仁兄，看题了吗？特别是问题补充~~

Answer 3

解：（x-2)(x+3)+4
=x^2-x-6+4
=x^2-x-2
=(x-2)(x+1)

Answer 4

确定没写错题吗 貌似在初中范围内 这个是分解不了的哦 再检查下吧