中二數學(~急~)
多項式簡介
分解為因式
1.xyp-xyp+3p-3p
回答 (13)
✔ 最佳答案
1.xyp-xyp+3p-3p
=p(xy-xy+3-3)
=p(0+0)
=p(0)
=0
參考: me
你有幾個計算方法︰
1.xyp-xyp+3p-3p
=xyp(1-1)+3p(1-1)
=xyp(0)+3p(0) ~0乘任何數都係0
=0
2. xyp-xyp+3p-3p
=p(xy-xy+3p-3p)
=p(0)
=0
3.xyp-xyp+3p-3p ~xyp-xyp=0 ; 3p-3p=0 , so
=0
2007-11-03 20:34:33 補充:
更正第二題︰2 .xyp-xyp+3p-3p= p(xy-xy+3-3)= p(0)= 0not上面的第二題
參考: me
xyp-xyp+3p-3p
=xyp-xyp+3p-3p (對消)
=0
xyp-xyp+3p-3p
=xyp-3p-xyp+3p
=p(xy-3)-p(xy-3)
=(xy-3)(p-p)
=0(xy-3)
要識多項式,先要識單項式
單項式即是一個代數式或一個數字或一個變數
例如:1,-5,0.3ab.........
多項式即是由兩個單項式相加而成
例如:y+5,3k.......
在多項式中,每個單項式稱為該多項式的項,而只含數字的項稱為常數項,一個多項式中最高次項的數,便是該多項式的次數
例如:3k-9ac+4是一個二次多項式,常數項:4
因式分解:
1.xyp-xyp+3p-3p
=p(xy-xy+3-3)
=p(xy-xy)
=p(0)
=0
參考: maths book
xyp(1-1)+3p(1-1)
=xyp(0)+3p(0)
=0
xyp-xyp+3p-3p
=p(xy-xy+3-3)
=0
xyp-xyp+3p-3p
=p(xy-xy+3-3)
1.xyp-xyp+3p-3p
=p(xy-xy+3-3)
=p(0+0)
=p(0)
=0
參考: my
收錄日期: 2021-04-13 19:15:57
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