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標題:

數學題(急急急!!!)

發問:

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~

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最佳解答:

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.

其他解答:

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