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18. (本小题7分)若$|a + b - 6|+(ab - 4)^{2}=0$,求$-a^{3}b - 2a^{2}b^{2}-ab^{3}$的值.
答案：
解：$\because|a + b - 6|+(ab - 4)^{2}=0$,
$\therefore a + b - 6 = 0$且$ab - 4 = 0$, 则$a + b = 6$, $ab = 4$.
$\therefore -a^{3}b-2a^{2}b^{2}-ab^{3}=-ab(a^{2}+2ab + b^{2})=-ab(a + b)^{2}=-4\times6^{2}=-144$. 即：$-a^{3}b-2a^{2}b^{2}-ab^{3}=-144$.
$\therefore a + b - 6 = 0$且$ab - 4 = 0$, 则$a + b = 6$, $ab = 4$.
$\therefore -a^{3}b-2a^{2}b^{2}-ab^{3}=-ab(a^{2}+2ab + b^{2})=-ab(a + b)^{2}=-4\times6^{2}=-144$. 即：$-a^{3}b-2a^{2}b^{2}-ab^{3}=-144$.
19. (本小题8分)如图,在一个大圆盘中,镶嵌着四个大小一样的小圆盘,已知大小圆盘的半径都是整数,阴影部分的面积为$5\pi\ cm^{2}$,请你求出大小两个圆盘的半径.
答案：
解：设大圆盘的半径为$R$ cm,
一个小圆盘的半径为$r$ cm, 根据题意，
得：$\pi R^{2}-4\pi r^{2}=5\pi$, 即$(R + 2r)(R - 2r)=5$.
$\because R$, $r$均为正整数,$\therefore R + 2r$, $R - 2r$也为正整数,
$\therefore\begin{cases}R + 2r = 5\\R - 2r = 1\end{cases}$, 解得$\begin{cases}R = 3\\r = 1\end{cases}$.
答：大圆盘的半径为 3 cm, 一个小圆盘的半径为 1 cm.
一个小圆盘的半径为$r$ cm, 根据题意，
得：$\pi R^{2}-4\pi r^{2}=5\pi$, 即$(R + 2r)(R - 2r)=5$.
$\because R$, $r$均为正整数,$\therefore R + 2r$, $R - 2r$也为正整数,
$\therefore\begin{cases}R + 2r = 5\\R - 2r = 1\end{cases}$, 解得$\begin{cases}R = 3\\r = 1\end{cases}$.
答：大圆盘的半径为 3 cm, 一个小圆盘的半径为 1 cm.
20. (本小题8分)两位同学将一个关于$x$的二次三项式$ax^{2}+bx + c$分解因式时,一位同学因看错了一次项系数而分解成$2(x - 1)(x - 9)$,另一位同学因看错了常数项而分解成$2(x - 2)(x - 4)$.请求出$a,b,c$的值,并将这个二次三项式因式分解.
答案：
解：$2(x - 1)(x - 9)$
$=2(x^{2}-x - 9x + 9)$
$=2(x^{2}-10x + 9)$
$=2x^{2}-20x + 18$,
$\because$一位同学因看错了一次项系数而分解成$2(x - 1)(x - 9)$, $\therefore a = 2$, $c = 18$;
$2(x - 2)(x - 4)$
$=2(x^{2}-2x - 4x + 8)$
$=2(x^{2}-6x + 8)$
$=2x^{2}-12x + 16$,
$\because$另一位同学因看错了常数项而分解成$2(x - 2)(x - 4)$, $\therefore b = -12$,
$\therefore$原多项式为$2x^{2}-12x + 18$,
$\therefore 2x^{2}-12x + 18$
$=2(x^{2}-6x + 9)$
$=2(x - 3)^{2}$
$=2(x^{2}-x - 9x + 9)$
$=2(x^{2}-10x + 9)$
$=2x^{2}-20x + 18$,
$\because$一位同学因看错了一次项系数而分解成$2(x - 1)(x - 9)$, $\therefore a = 2$, $c = 18$;
$2(x - 2)(x - 4)$
$=2(x^{2}-2x - 4x + 8)$
$=2(x^{2}-6x + 8)$
$=2x^{2}-12x + 16$,
$\because$另一位同学因看错了常数项而分解成$2(x - 2)(x - 4)$, $\therefore b = -12$,
$\therefore$原多项式为$2x^{2}-12x + 18$,
$\therefore 2x^{2}-12x + 18$
$=2(x^{2}-6x + 9)$
$=2(x - 3)^{2}$
21. (本小题12分)阅读材料:若$m^{2}-2mn + 2n^{2}-8n + 16 = 0$,求$m,n$的值.
解:$\because m^{2}-2mn + 2n^{2}-8n + 16 = 0$,
$\therefore (m^{2}-2mn + n^{2})+(n^{2}-8n + 16)=0$,
$\therefore (m - n)^{2}+(n - 4)^{2}=0$,
$\therefore (m - n)^{2}=0,(n - 4)^{2}=0$,
$\therefore n = 4,m = 4$.
根据你的观察,探究下面的问题:
(1)若$a^{2}+b^{2}-4a + 4 = 0$,则$a =$________,$b =$________;
(2)已知$x^{2}+2y^{2}-2xy + 6y + 9 = 0$,求$x^{y}$的值;
(3)已知$\triangle ABC$的三边长$a,b,c$都是正整数,且满足$2a^{2}+b^{2}-4a - 6b + 11 = 0$,求$\triangle ABC$的周长.
解:$\because m^{2}-2mn + 2n^{2}-8n + 16 = 0$,
$\therefore (m^{2}-2mn + n^{2})+(n^{2}-8n + 16)=0$,
$\therefore (m - n)^{2}+(n - 4)^{2}=0$,
$\therefore (m - n)^{2}=0,(n - 4)^{2}=0$,
$\therefore n = 4,m = 4$.
根据你的观察,探究下面的问题:
(1)若$a^{2}+b^{2}-4a + 4 = 0$,则$a =$________,$b =$________;
(2)已知$x^{2}+2y^{2}-2xy + 6y + 9 = 0$,求$x^{y}$的值;
(3)已知$\triangle ABC$的三边长$a,b,c$都是正整数,且满足$2a^{2}+b^{2}-4a - 6b + 11 = 0$,求$\triangle ABC$的周长.
答案：
解：
(1)2 0
(2)$\because x^{2}+2y^{2}-2xy + 6y + 9 = 0$,
$\therefore x^{2}+y^{2}-2xy + y^{2}+6y + 9 = 0$,
即$(x - y)^{2}+(y + 3)^{2}=0$, 则$x - y = 0$, $y + 3 = 0$,
解得$x = y = -3$, $\therefore x^{y}=(-3)^{-3}=-\frac{1}{27}$.
(3)$\because 2a^{2}+b^{2}-4a - 6b + 11 = 0$,
$\therefore 2a^{2}-4a + 2 + b^{2}-6b + 9 = 0$,
$\therefore 2(a - 1)^{2}+(b - 3)^{2}=0$, 则$a - 1 = 0$, $b - 3 = 0$,
解得$a = 1$, $b = 3$, $\because a$, $b$, $c$都是正整数, 由三角形三边关系可知, 三角形的三边长分别为 1, 3, 3, 则$\triangle ABC$的周长为$1 + 3 + 3 = 7$.
(1)2 0
(2)$\because x^{2}+2y^{2}-2xy + 6y + 9 = 0$,
$\therefore x^{2}+y^{2}-2xy + y^{2}+6y + 9 = 0$,
即$(x - y)^{2}+(y + 3)^{2}=0$, 则$x - y = 0$, $y + 3 = 0$,
解得$x = y = -3$, $\therefore x^{y}=(-3)^{-3}=-\frac{1}{27}$.
(3)$\because 2a^{2}+b^{2}-4a - 6b + 11 = 0$,
$\therefore 2a^{2}-4a + 2 + b^{2}-6b + 9 = 0$,
$\therefore 2(a - 1)^{2}+(b - 3)^{2}=0$, 则$a - 1 = 0$, $b - 3 = 0$,
解得$a = 1$, $b = 3$, $\because a$, $b$, $c$都是正整数, 由三角形三边关系可知, 三角形的三边长分别为 1, 3, 3, 则$\triangle ABC$的周长为$1 + 3 + 3 = 7$.
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