1.展开(y+3z)(y-3z)
2.因式分解p 2次r+pr-q2次r-qr
展开及因式分解
2012-01-03 3:03 am
回答 (5)
2012-01-05 3:22 am
✔ 最佳答案
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)
參考: Hope I Can Help You ^_^ ( From me )
2012-01-05 3:42 am
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)
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)
2012-01-04 9:46 pm
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)
(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)
2012-01-04 4:32 am
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。
(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。
參考: 俺自己
2012-01-03 3:11 am
(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)
第二條系唔系呢個樣？p^2(r+pr)-q^2(r-qr)
參考: myself
收錄日期: 2021-04-20 12:03:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120102000051KK00796