數學題(急急急!!!)

1.x^2-(y-z)^2
2.(x+y+z)(x+y+z)
3.[2x+(3y-2)][2x-(3y-2)]
4 證明(x+y)^3≡x^3+3x^2y+3xy^2+y^3

thx~

回答 (2)

2007-07-17 6:02 pm
✔ 最佳答案
x2 – (y - z)2
= [x + (y - z)][x – (y - z)]
= (x + y - z)(x – y + z)

(x + y + z)(x + y + z)
= x(x + y + z) + y(x + y + z) + z(x + y + z)
= x2 + xy + xz + xy + y2 yz + xz + yz + z2
= x2 + y2 + z2 + 2xy + 2yz + 2xz

[2x + (3y - 2)][2x – (3y - 2)]
= (2x)2 – (3y - 2)2
= 4x2 – (9y2 – 12y + 4)
= 4x2 – 9y2 + 12y – 4

證明(x + y)3 ≡ x3 + 3x2y + 3xy2 + y3

L.H.S.
= (x + y)3
= (x + y)(x + y)(x + y)
= [x(x + y) + y(x + y)](x + y)
= (x2 + 2xy + y2)(x + y)
= x2(x + y) + 2xy(x + y) + y2(x + y)
= x3 + x2y + 2x2y + 2xy2 + xy2 + y3
= x3 + 3x2y + 3xy2 + y3
= R.H.S.
參考: Myself~~~
2007-07-17 6:11 pm
as A^2 - B^2 = (A+B)(A-B)

1. x^2-(y-z)^2
= [x-(y-z)][x+(y-z)]
= (x-y+z)(x+y-z)

2. (x+y+z)(x+y+z)
= [(x+y)+z]^2
= (x+y)^2 + 2(x+y)z + z^2
= x^2 + y^2 + 2xy + 2xz + 2yz + z^2
= x^2 + y^2 + z^2 + 2(xy + xz + yz)

3. [2x+(3y-2)][2x-(3y-2)]
= (2x)^2 - (3y-2)^2
= 4x^2 - (9y^2 -12y + 4)
= 4x^2 - 9y^2 + 12y - 4

4. LHS = (x+y)^3
(x+y)^3 = (x+y)(x+y)(x+y)
= (x^2 + 2xy + y^2)(x+y)
= x^3 + 2x^2y+ xy^2 + x^2y + 2xy^2 + y^3
= x^3 + 2x^2y+ x^2y + xy^2 + 2xy^2 + y^3 (調左位, 易睇 d, 第2&3 項, 第4&5項 係一樣)
= x^3 + 3x^2y + 3xy^2 +y^3
= RHS


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