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第77页

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$x\neq-\frac{1}{2}$

$-1$或$3$或$1$

 解：$\begin{aligned}&\frac{1}{3}a^{2}b^{3}\cdot(3ab^{2})^{2}\div6(a^{2}b^{3})^{2}\\=&\frac{1}{3}a^{2}b^{3}\cdot9a^{2}b^{4}\div6a^{4}b^{6}\\=&( \frac{1}{3}\times9\div6)\times(a^{2 + 2-4})\times(b^{3 + 4-6})\\=&\frac{1}{2}b\end{aligned}$

 解：$\begin{aligned}&(-52x^{n + 4}+20x^{n + 2}-4x^{n - 2})\div(-4x^{n - 2})\\=&(-52x^{n + 4})\div(-4x^{n - 2})+20x^{n + 2}\div(-4x^{n - 2})-4x^{n - 2}\div(-4x^{n - 2})\\=&13x^{(n + 4)-(n - 2)}-5x^{(n + 2)-(n - 2)} + 1\\=&13x^{6}-5x^{4}+1\end{aligned}$

 解：$\begin{aligned}&[x(x^{2}y^{2}-xy)-y(x^{2}-x^{3}y)]\div(-3x^{2}y)\\=&(x^{3}y^{2}-x^{2}y - x^{2}y+x^{3}y^{2})\div(-3x^{2}y)\\=&(2x^{3}y^{2}-2x^{2}y)\div(-3x^{2}y)\\=&2x^{3}y^{2}\div(-3x^{2}y)-2x^{2}y\div(-3x^{2}y)\\=&-\frac{2}{3}xy+\frac{2}{3}\end{aligned}$

 解：

 $\begin{aligned}&(3a^{2}b - ab^{2}-2b^{3})\div b-(a - 2b)(b - 2a)\\=&3a^{2}-ab - 2b^{2}-(ab-2a^{2}-2b^{2}+4ab)\\=&3a^{2}-ab - 2b^{2}-ab + 2a^{2}+2b^{2}-4ab\\=&5a^{2}-6ab\end{aligned}$

 因为$(a - 1)^{2}+\vert b + 1\vert = 0，$又因为$(a - 1)^{2}\geq0，$$\vert b + 1\vert\geq0，$所以$a - 1 = 0，$$b + 1 = 0，$即$a = 1，$$b = - 1。$

 当$a = 1，$$b = - 1$时，原式$=5\times1^{2}-6\times1\times(-1)=5 + 6 = 11。$

 解：

 $\begin{aligned}(y^{2})^{m}\cdot(x^{n + 1})^{2}\div(x^{n}y)&=x^{3}y^{3}\\y^{2m}\cdot x^{2n + 2}\div(x^{n}y)&=x^{3}y^{3}\\x^{2n + 2 - n}y^{2m - 1}&=x^{3}y^{3}\\x^{n + 2}y^{2m - 1}&=x^{3}y^{3}\end{aligned}$

 所以$2m - 1 = 3，$$n + 2 = 3。$

 由$2m - 1 = 3，$得$2m = 4，$$m = 2；$由$n + 2 = 3，$得$n = 1。$

 解：(1) 若多项式有因式$x - 2，$则此多项式能被$x - 2$整除；若$x - 2 = 0，$则此多项式的值为$0。$

 (2) $M$能被$x - k$整除。

 (3) 因为$x - 2$能整除$x^{2}+kx - 14，$所以当$x - 2 = 0$时，$x^{2}+kx - 14 = 0，$

即当$x = 2$时，$x^{2}+kx - 14 = 4 + 2k - 14 = 0，$解得$k = 5。$