求數學高手幫忙

2006-10-15 12:03 am
Q1)x²y²+xy-x²-y²+x+y+2

Q2)1-12x²y²+48(x²y²)²-64(x²y²)³

回答 (1)

2006-10-15 12:36 am
✔ 最佳答案
1. x^2 y^2 + xy - x^2 - y^2 + x + y + 2
= (x^2 y^2 - x^2 - y^2 + 1) + (xy + x + y + 1)
= x^2(y^2 - 1) - (y^2 - 1) + x(y + 1) + (y + 1)
= (x^2 - 1)(y^2 - 1) + (x + 1)(y + 1)
= (x + 1)(x - 1)(y + 1)(y - 1) + (x + 1)(y + 1)
= (x + 1)(y + 1)[(x - 1)(y - 1) + 1]
= (x + 1)(y + 1)(xy - x - y + 1 - 1)
= (x + 1)(y + 1)(xy - x - y)
2. 1 - 12x^2 y^2 + 48(x^2 y^2)^2 - 64(x^2 y^2)^3
= 1 - 64(x^2 y^2)^3 - 12x^2 y^2(1 - 4x^2 y^2)
= {1 - [4(x^2 y^2)]^3} - 12x^2 y^2(1 - 4x^2 y^2)
= (1 - 4(x^2 y^2))[1 + 4(x^2 y^2) + 16(x^2 y^2)^2] - 12x^2 y^2(1 - 4x^2 y^2)
= (1 - 4(x^2 y^2))[1 + 4(x^2 y^2) + 16(x^2 y^2)^2 - 12x^2 y^2]
= (1 - 4(x^2 y^2))[1 - 8(x^2 y^2) + 16(x^2 y^2)^2]
= (1 - 4(x^2)y^2)(1 - 4(x^2)y^2)^2
= (1 - 4(x^2)y^2)^3
= [(1 + 2xy)(1 - 2xy)]^3
= (1 + 2xy)^3 * (1 - 2xy)^3

2006-10-14 16:38:11 補充:
樓上那位仁兄: 我們要做的是因式分解......


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