扫描以下二维码下载并安装猿辅导App, 打开后请搜索教师姓名"赵胤"即可报名本课程(14次课, 99元).
分解下列因式:
1. $x^3 + 3xy + y^3 - 1$
解答: $$x^3 + 3xy + y^3 - 1 = x^3 + 3xy(x + y) + y^3 - 1 - 3xy(x + y) + 3xy$$ $$= (x + y)^3 - 1 - 3xy(x + y - 1)$$ $$= (x + y - 1)(x^2 + 2xy + y^2 + x + y + 1) - 3xy(x + y - 1)$$ $$= (x + y - 1)(x^2 + y^2 - xy + x + y + 1).$$ 另解: $$x^3 + 3xy + y^3 - 1 = x^3 + y^3 + (-1)^3 - 3xycdot(-1)$$ $$ = (x + y - 1)(x^2 + y^2 + 1 - xy + x + y).$$
2. $x^4 + 4$
解答: $$x^4 + 4 = x^4 + 4x^2 + 4 - 4x^2$$ $$= (x^2 + 2)^2 - 4x^2$$ $$= (x^2 + 2 + 2x)(x^2 + 2 - 2x).$$
3. $a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2$
解答: $$a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2 = ab(a - b) + bc(b - c) + ca(c - a)$$ $$= ab(a - b) - bc[(a - b) + (c - a)] + ca(c - a)$$ $$= (a - b)(ab - bc) + (c - a)(ca - bc)$$ $$= (a - b)(c - a)(c - b).$$ 另解: $$a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2 = ab(a - b) + c^2(a - b) - c(a^2 - b^2) $$ $$= (a - b)(ab - ca - cb + c^2) = (a - b)(b - c)(a - c).$$
4. $x^2 + y^2 - x^2y^2 - 4xy - 1$
解答: $$x^2 + y^2 - x^2y^2 - 4xy - 1 = (x^2 + y^2 - 2xy) - (x^2y^2 + 2xy + 1)$$ $$= (x - y)^2 - (xy + 1)^2 = (x - y + xy + 1)(x - y - xy - 1).$$
5. $x^3 - x^2 - x - 2$
解答: $$x^3 - x^2 - x - 2 = (x^3 - 1) - (x^2 + x + 1)$$ $$= (x - 1)(x^2 + x + 1) - (x^2 + x + 1)= (x^2 + x + 1)(x - 2).$$
6. $a^4 - b^4 + c^4 - d^4 - 2(a^2c^2 - b^2d^2) + 4ac(b^2 + d^2) - 4bd(a^2 + c^2)$
解答: $$a^4 - b^4 + c^4 - d^4 - 2(a^2c^2 - b^2d^2) + 4ac(b^2 + d^2) - 4bd(a^2 + c^2)$$ $$= (a^4 + c^4 - 2a^2c^2) - (b^4 + d^4 - 2b^2d^2) + 4ac(b^2 + d^2) - 4bd(a^2 + c^2)$$ $$= (a^2 - c^2)^2 - (b^2 - d^2)^2 + 4ac(b^2 + d^2) - 4bd(a^2 + c^2)$$ $$= (a+c)^2(a - c)^2 - (b + d)^2(b - d)^2 + 4ac[(b + d)^2 - 2bd] - 4bd[(a + c)^2 - 2ac]$$ $$= (a+c)^2(a - c)^2 - (b + d)^2(b - d)^2 + 4ac(b + d)^2 - 4bd(a + c)^2$$ $$= (a+c)^2(a - c)^2 - (b + d)^2(b - d)^2 + [(a + c)^2 - (a - c)^2](b + d)^2 - [(b + d)^2 - (b - d)^2](a + c)^2$$ $$= (a - c)^2[(a + c)^2 - (b + d)^2] + (b - d)^2[(a + c)^2 - (b + d)^2]$$ $$= [(a + c)^2 - (b + d)^2][(a - c)^2 + (b - d)^2]$$ $$= (a + c + b + d)(a + c - b - d)(a^2 + b^2 + c^2 + d^2 - 2ac - 2bd).$$ 评注: 本题使用了以下事实 $$4xy = (x + y)^2 - (x - y)^2.$$
7. $(a-b)^3 + (b-a-2)^3 + 8$
解答: $$(a-b)^3 + (b-a-2)^3 + 8 = (a-b)^3 + (b-a-2)^3 + 2^3,$$ 而$(a - b) + (b - a - 2) + 2 = 0$, 因此 $$(a-b)^3 + (b-a-2)^3 + 8 = 3(a - b)(b - a - 2)cdot2 = 6(a - b)(b - a - 2).$$
8. 求证: 如果一个数可以表示成两个整数的平方和, 那么这个数的2倍也可以表示成两个整数的平方和.
解答:
设 $x = m^2 + n^2$, ($m, ninmathbf{Z}$). 则 $$2x = 2m^2 + 2n^2 = (m^2 + n^2 + 2mn) + (m^2 + n^2 - 2mn) = (m + n)^2 + (m - n)^2.$$ 证毕.
9. 求能使$m^2 + m + 7$是完全平方数的所有整数$m$的积是多少?
解答:
设 $m^2 + m + 7 = x^2$. 则 $$(m+{1over2})^2 - x^2 = -{27over4}$$ $$Rightarrow 4x^2 - (2m+1)^2 = 27$$ $$Rightarrow (2x + 2m + 1)(2x - 2m - 1) = 27$$ $$Rightarrow egin{cases}2x + 2m + 1 = 1, -1, 27, -27, 3, -3, 9, -9\ 2x - 2m - 1 = 27, -27, 1, -1, 9, -9, 3, -3 end{cases}$$ $$Rightarrow m = -7, 6, -2, 1.$$ 故其乘积为 $-7 imes6 imes-2 imes1 = 84$.
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