展开及因式分解

您好，我是 lop，高興能解答您的問題。

(1)

(y+3z)(y-3z)
= y(y-3z) + 3z(y-3z)
= y²- 3yz + 3yz - 9z²
= y²- 9z²

(2)

p²r + pr - q²r - qr
= r(p²+ p - q²- q)
= r [ (p²+ pq + p) - (q²+ pq + q) ]
= r[ p(p+q+1) - q(p+q+1) ]
= r(p-q)(p+q+1)

1,(y+3z)(y-3z)

a^2-b^2=(a+b)(a-b)

solution:
y^2 - 3^2 z^2
=y^2 - 9z^2

2.因式分解p 2次r+pr-q2次r-qr

p^2r + pr - q^2 r- qr

solution:

p^2r + pr - q^2 r- qr

=r(p^2 + p - q^2 - q)

=r(p^2- q^2) + p - q

=r(p-2)^2+p-g--(p^2- q^2)
a^2-b^2=(a+b)(a-b)
so.....
=r(p+q)(p-q)+p-q

=r(p-q)(p+q+1)

1.
(y+3z)(y-3z)
=y^2-9z^2

2.
p^2r+pr-q^2r-qr
=p^2r-q^2r+pr-qr
=r(p^2-q^2)+r(p-q)
=r[(p^2-q^2)+(p-q)]
=r[(p-q)(p+q)+(p-q)]
=r(p-q)[(p+q)+1]
=r(p-q)(p+q+1)

1.展开(y+3z)(y-3z)
(y+3z)(y-3z)
=(y^2)-3z．y+3y．z-3z．3z
=(y^2)+3yz-3yz-(9z^2)
=y^2-9z^2#

2.因式分解p 2次r+pr-q2次r-qr
p^2r+pr-q^2r-qr
=r(p^2+p-q^2-q)...我相信這像是答案=.=
=r(p^3-q)#...純屬個人見解。

(a+b)(a-b)=a^2-b^2(y+3z)(y-3z)=(y)^2-(3z)^2=y^2-(3^2)(z^2)=y^2-9z^2

第二條系唔系呢個樣？p^2(r+pr)-q^2(r-qr)