Question

计算整式

（x^2-2xy+4y^2)(x+2y)-8x(x+y)(x-y)
(64a^3-27b^3)\(4a-3b)-(4a+3b)^2
(x^4-x+1）/（x^2+2x-3)

Answer 1

又做到你的题了

1）(x^2-2xy+4y^2)(x+2y)-8x(x+y)(x-y)
立方和公式：(a+b)(a²-ab+b²)=a³+b³
(x^2-2xy+4y^2)(x+2y)
=(x+2y)[x²-x×2y+(2y)²]
=x³+8y³

8x(x+y)(x-y)
=8x[(x+y)(x-y)]
=8x(x²-y²)
=8x³-8xy²

所以(x^2-2xy+4y^2)(x+2y)-8x(x+y)(x-y)
=x³+8y³-(8x³-8xy²)
= x³+8y³-8x³+8xy²
=-7x³+8xy²+8y³

2）(64a^3-27b^3)\(4a-3b)-(4a+3b)^2
立方差公式：a³-b³=(a-b)(a²+ab+b²)

(64a^3-27b^3)\(4a-3b)
=[(4a)³-(3b)³]/(4a-3b)
=(4a-3b)[(4a)²+4a×3b+(3b)²]/ (4a-3b)
=[(4a)²+2×4a×3b+(3b)²]-12ab
=(4a+3b)²-12ab

所以(64a^3-27b^3)\(4a-3b)-(4a+3b)^2
=(4a+3b)²-12ab-(4a+3b)²
=-12ab

3）(x^4-x+1）/（x^2+2x-3)
题目有没有出错呢？
还是按原来的做
x^4-x+1
=(x^4-2x+1)+x
=[(x^4-2x²+1)+(2x²-2x)]+x
=[(x²-1)²+2x(x-1)]+(x-1)+1
=[(x+1)²(x-1)²+2x(x-1)+(x-1)]+1
=(x-1)[(x+1)²(x-1)+2x+1]+1
=(x-1)(x³-x+x²-1+2x+1)+1
=(x-1)(x³+x²+x)+1

（x^2+2x-3)
=(x+3)(x-1)
所以(x^4-x+1）/（x^2+2x-3)
=【(x-1)(x³+x²+x)+1】/(x+3)(x-1)
=( x³+x²+x )/(x+3)+[1/(x^2+2x-3)]
=(x³+3x²-2x²-6x+7x+21-21) /(x+3)+[1/(x^2+2x-3)]
=x²-2x+7-[21/(x+2)] +[1/(x^2+2x-3)]

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