代數式MATHS ,HELP!!!

2010-09-27 6:02 am
展開下列代數式:
X(A+B)+Y(A+B)

2A(X+1)+3B(X+1)

(A+1)+(A+1)

X(B-A)+Y(A-B)

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因式分解下列代數式:
3(5-2a)-x(2a-5)

X-Y-B(Y-X)

X(B-A)+y(A-B)

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回答 (2)

2010-09-27 8:23 am
✔ 最佳答案
展開下列代數式:

X(A+B)+Y(A+B) = XA +XB +YA +YB

2A(X+1)+3B(X+1) = 2AX +2A + 3BX + 3B

(A+1)+(A+1) = A + 1 + A + 1 = 2A + 2

X(B-A)+Y(A-B) = XB –XA +YA - YB

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因式分解下列代數式:

3(5-2a)-x(2a-5) = 3(5 – 2a) +x(-2a + 5) = 3(5 – 2a) +x(5 -2a) = (5-2a) (3 + x)

X-Y-B(Y-X) = 1(X- Y) –B(Y – X) =1(X- Y) +B(-Y + X) =1(X- Y) +B (X -Y)
= (X – Y) (1 +B)

X(B-A)+Y(A-B) = X(B-A)-Y(-A+B) = X(B-A)-Y(B-A) = (B-A) (X-Y)
2010-09-27 6:16 am
展開下列代數式:
X(A+B)+Y(A+B)
= AX+BX+AY+BY

2A(X+1)+3B(X+1)
= 2AX+2A+3BX+3B

(A+1)+(A+1)
= A+1+A+1
= 2A+2 我唔肯定…

X(B-A)+Y(A-B)
= BX-AX+AY-BY


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