求證以下是恆等式(詳細展開公式)
(2x-1)(x-2)=x(2x-4)-(x-2)
恆等式一問
2006-11-03 12:10 am
回答 (4)
2006-11-03 12:48 am
Ans:
LHS:
(2x - 1)(x - 2)
RHS:
x(2x - 4) - (x - 2)
= 2x^2 - 4x - x + 2
= 2x^2 - 5x + 2
= (2x - 1)(x - 2) {By Factorization}
So, LHS = RHS
LHS:
(2x - 1)(x - 2)
RHS:
x(2x - 4) - (x - 2)
= 2x^2 - 4x - x + 2
= 2x^2 - 5x + 2
= (2x - 1)(x - 2) {By Factorization}
So, LHS = RHS
2006-11-03 12:27 am
(2x-1)(x-2)=x(2x-4)-(x-2)
LHS=(2x-1)(x-2)
=2X(X-2)-(X-2)
=2X^2-4X-X+2
=2X^2-5X+2
RHS=x(2x-4)-(x-2)
=2X^2-4X-X+2
=2x^2-5X+2=LHS
SO.(2x-1)(x-2)=x(2x-4)-(x-2) (注意＝係有三劃，not二劃）
LHS=(2x-1)(x-2)
=2X(X-2)-(X-2)
=2X^2-4X-X+2
=2X^2-5X+2
RHS=x(2x-4)-(x-2)
=2X^2-4X-X+2
=2x^2-5X+2=LHS
SO.(2x-1)(x-2)=x(2x-4)-(x-2) (注意＝係有三劃，not二劃）
2006-11-03 12:22 am
(2x-1)(x-2)=x(2x-4)-(x-2)
(2x-1)(x)-(2x-1)(2)=2x^2-4x-(x-2)
(2x^2-x)-(4x-2)=2x^2-4x-x+2
2x^2-x-4x+2=2x^2-5x+2
2x^2-5x+2=2x^2-5x+2
Because left=right,
So (2x-1)(x-2)=x(2x-4)-(x-2)是恆等式.
(2x-1)(x)-(2x-1)(2)=2x^2-4x-(x-2)
(2x^2-x)-(4x-2)=2x^2-4x-x+2
2x^2-x-4x+2=2x^2-5x+2
2x^2-5x+2=2x^2-5x+2
Because left=right,
So (2x-1)(x-2)=x(2x-4)-(x-2)是恆等式.
參考: me
2006-11-03 12:21 am
(2x-1)(x-2)=x(2x-4)-(x-2)
(LHS)(2x-1)(x-2)=2x^2-4x-x+2
=2x^2-5x+2
2006-11-02 16:30:53 補充：
(RHS)=2x^2 -4x-x 2=2x^2-5x 2therefor, (2x-1)(x-2)全等於x(2x-4)-(x-2).
(LHS)(2x-1)(x-2)=2x^2-4x-x+2
=2x^2-5x+2
2006-11-02 16:30:53 補充：
(RHS)=2x^2 -4x-x 2=2x^2-5x 2therefor, (2x-1)(x-2)全等於x(2x-4)-(x-2).
收錄日期: 2021-04-12 23:24:56
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