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標題:
中三數學問題求幫忙
發問:
1) 5x^3-20x(x+y)^2 2) 48xy^2-75xz^2 3) 4x^2-8xy+4y^2-9z^2 4) x^4-21x^2-100 5) x^3-(x+y)^3 6) 1-729x^2 7) Factorize x^8-1 and hence factorize x^1024-1. 8a) Factorize x^3-64. 8b) Hence factorize x^3-(64/x^3). 有睇吾明可以問我 知道答案的plz help.
最佳解答:
1) 5x^3-20x(x+y)^2 =5x[x^2-2^2(x+y)^2] =5x{[2(x+y)+x][x-2(x+y]} =5x(3x+2y)(-x-2y) =-5x (3x+2y) (x+2y) 2) 48xy^2-75xz^2 =3x(16y^2-25z^2) =3x(4y-5z)(4y+5z) 3) 4x^2-8xy+4y^2-9z^2 = 4(x^2 - 2xy + y^2) - 9z^2 = 4(x - y)^2 - 9z^2 = (2(x-y) + 3z) (2(x-y) - 3z) = (2x - 2y + 3z)( 2x - 2y - 3z) 4) x^4 - 21x^2 - 100 = x^2 * x^2 - 21x^2 - 100 = (x^2)^2 - 21x^2 - 100 = x^2 ( x^2 - 21) - 100 =x^2(x - sq root 21)(x + sq root 21) - 100 5) x^3 - (x + y)^3 = (x - (x - y)) (x^2 + x(x + y) + (x + y)^2) = -y(3x^2 + 2xy + y^2) 6) 1-729x^2 = 1 - 27^2 * x^2 = (1 - 27x)(1 + 27x) 7) x^8 - 1 = (x^4)^2 - 1 = (x^4 - 1)(x^4 + 1) Hence, x^1024 - 1 = (x^128)^8 - 1 = [(x^128)^4 - 1][(x^128)^4 + 1] = (x^512 - 1)(x^512 + 1) 8a) x^3 - 64 = x^3 - 4^3 = (x - 4)(x^2 + 4x + 16) b) Hence, x^3 - (64 / x^3) = x^3 - (4 / x)^3 (4 / x) can be regarded as 4 in part (a) = [x - (4/x)][x^2 + (4/x)(x) + 16 / x^2] = [x - (4/x)][x^2 + 4 + 16 / x^2]
其他解答:DF665233EBFAA7EC
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