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1. 若关于x,y的方程组$\left\{\begin{array}{l} x+ay+1=0,\\ bx-2y+a=0\end{array}\right. $没有实数解,则 （ ）
A. $ab=-2$
B. $ab=-2$且$a≠1$
C. $ab≠-2$
D. $ab=-2$且$a≠2$
A. $ab=-2$
B. $ab=-2$且$a≠1$
C. $ab≠-2$
D. $ab=-2$且$a≠2$
答案：
A
方程组无解需$\frac{1}{b}=\frac{a}{-2}≠\frac{-1}{-a}$。由$\frac{1}{b}=\frac{a}{-2}$得$ab=-2$，此时$\frac{a}{-2}=-\frac{a}{2}$，$\frac{-1}{-a}=\frac{1}{a}$，若$ab=-2$，则$-\frac{a}{2}≠\frac{1}{a}$即$a^2≠-2$恒成立，故只需$ab=-2$。
方程组无解需$\frac{1}{b}=\frac{a}{-2}≠\frac{-1}{-a}$。由$\frac{1}{b}=\frac{a}{-2}$得$ab=-2$，此时$\frac{a}{-2}=-\frac{a}{2}$，$\frac{-1}{-a}=\frac{1}{a}$，若$ab=-2$，则$-\frac{a}{2}≠\frac{1}{a}$即$a^2≠-2$恒成立，故只需$ab=-2$。
2. $\frac {x+y}{2}=\frac {2x-y}{3}=x+2$的解是______.
答案：
$\left\{\begin{array}{l} x=-5\\ y=-1\end{array}\right. $
设$\frac{x+y}{2}=\frac{2x-y}{3}=k=x+2$，则$x=k-2$，$x+y=2k$得$y=k+2$，$2x-y=3k$得$2(k-2)-(k+2)=3k$，解得k=-3，故x=-5，y=-1。
设$\frac{x+y}{2}=\frac{2x-y}{3}=k=x+2$，则$x=k-2$，$x+y=2k$得$y=k+2$，$2x-y=3k$得$2(k-2)-(k+2)=3k$，解得k=-3，故x=-5，y=-1。
3. 若$2x-3y+z=0,3x-2y-6z=0$且$xyz≠0$,则$\frac {x^{2}+y^{2}+z^{2}}{2x^{2}+y^{2}-z^{2}}=$______.
答案：
$\frac{13}{20}$
由$2x-3y=-z$，$3x-2y=6z$，解得$x=4z$，$y=3z$。代入原式得$\frac{(16z^2)+(9z^2)+(z^2)}{2(16z^2)+(9z^2)-(z^2)}=\frac{26z^2}{40z^2}=\frac{13}{20}$。
由$2x-3y=-z$，$3x-2y=6z$，解得$x=4z$，$y=3z$。代入原式得$\frac{(16z^2)+(9z^2)+(z^2)}{2(16z^2)+(9z^2)-(z^2)}=\frac{26z^2}{40z^2}=\frac{13}{20}$。
4. 解下列方程组：
(1)$\left\{\begin{array}{l} |x|+|y|=7,\\ 2|x|-3|y|=-1;\end{array}\right. $
(2)$\left\{\begin{array}{l} 1995x+1997y=5989,\\ 1997x+1995y=5987;\end{array}\right. $
(3)$\left\{\begin{array}{l} \frac {pq}{p+q}=\frac {6}{5},\\ \frac {qr}{q+r}=\frac {3}{4},\\ \frac {rp}{r+p}=\frac {2}{3}.\end{array}\right. $
(1)$\left\{\begin{array}{l} |x|+|y|=7,\\ 2|x|-3|y|=-1;\end{array}\right. $
(2)$\left\{\begin{array}{l} 1995x+1997y=5989,\\ 1997x+1995y=5987;\end{array}\right. $
(3)$\left\{\begin{array}{l} \frac {pq}{p+q}=\frac {6}{5},\\ \frac {qr}{q+r}=\frac {3}{4},\\ \frac {rp}{r+p}=\frac {2}{3}.\end{array}\right. $
答案：
(1)$\left\{\begin{array}{l} x=4\\ y=3\end{array}\right. $，$\left\{\begin{array}{l} x=4\\ y=-3\end{array}\right. $，$\left\{\begin{array}{l} x=-4\\ y=3\end{array}\right. $，$\left\{\begin{array}{l} x=-4\\ y=-3\end{array}\right. $
设$a=|x|$，$b=|y|$，则$\left\{\begin{array}{l} a+b=7\\ 2a-3b=-1\end{array}\right. $，解得$a=4$，$b=3$，故$|x|=4$，$|y|=3$。
(2)$\left\{\begin{array}{l} x=1\\ y=2\end{array}\right. $
两式相加得$3992x+3992y=11976$即$x+y=3$；两式相减得$2x-2y=-2$即$x-y=-1$，解得$x=1$，$y=2$。
(3)$\left\{\begin{array}{l} p=2\\ q=3\\ r=1\end{array}\right. $
取倒数得$\left\{\begin{array}{l} \frac{1}{p}+\frac{1}{q}=\frac{5}{6}\\ \frac{1}{q}+\frac{1}{r}=\frac{4}{3}\\ \frac{1}{r}+\frac{1}{p}=\frac{3}{2}\end{array}\right. $，解得$\frac{1}{p}=\frac{1}{2}$，$\frac{1}{q}=\frac{1}{3}$，$\frac{1}{r}=1$，即$p=2$，$q=3$，$r=1$。
(1)$\left\{\begin{array}{l} x=4\\ y=3\end{array}\right. $，$\left\{\begin{array}{l} x=4\\ y=-3\end{array}\right. $，$\left\{\begin{array}{l} x=-4\\ y=3\end{array}\right. $，$\left\{\begin{array}{l} x=-4\\ y=-3\end{array}\right. $
设$a=|x|$，$b=|y|$，则$\left\{\begin{array}{l} a+b=7\\ 2a-3b=-1\end{array}\right. $，解得$a=4$，$b=3$，故$|x|=4$，$|y|=3$。
(2)$\left\{\begin{array}{l} x=1\\ y=2\end{array}\right. $
两式相加得$3992x+3992y=11976$即$x+y=3$；两式相减得$2x-2y=-2$即$x-y=-1$，解得$x=1$，$y=2$。
(3)$\left\{\begin{array}{l} p=2\\ q=3\\ r=1\end{array}\right. $
取倒数得$\left\{\begin{array}{l} \frac{1}{p}+\frac{1}{q}=\frac{5}{6}\\ \frac{1}{q}+\frac{1}{r}=\frac{4}{3}\\ \frac{1}{r}+\frac{1}{p}=\frac{3}{2}\end{array}\right. $，解得$\frac{1}{p}=\frac{1}{2}$，$\frac{1}{q}=\frac{1}{3}$，$\frac{1}{r}=1$，即$p=2$，$q=3$，$r=1$。
5. 已知方程组$\left\{\begin{array}{l} (m-n)x-3y=10①,\\ 4x+(3m+n)y=12②,\end{array}\right. $将$①×2-②$能消x,将$②+①$能消y,则m,n的值为多少?
答案：
$m=\frac{5}{4}$，$n=-\frac{3}{4}$
①×2-②得$[2(m-n)-4]x-[6+3m+n]y=8$，消x则$2(m-n)-4=0$即$m-n=2$；②+①得$[(m-n)+4]x+[-3+3m+n]y=22$，消y则$-3+3m+n=0$即$3m+n=3$。联立解得$m=\frac{5}{4}$，$n=-\frac{3}{4}$。
①×2-②得$[2(m-n)-4]x-[6+3m+n]y=8$，消x则$2(m-n)-4=0$即$m-n=2$；②+①得$[(m-n)+4]x+[-3+3m+n]y=22$，消y则$-3+3m+n=0$即$3m+n=3$。联立解得$m=\frac{5}{4}$，$n=-\frac{3}{4}$。
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