电子课本网 第66页

第66页

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$解:由题意,得关于x,y的方程组\begin{cases}{2x+y=-2​}\\ {ax+by=-4}\end{cases}​和\begin{cases}{3x-y=12​}\\ {bx+ay=-8}\end{cases}​的解也是方程组$
$\begin{cases}{ax+by=-4​}\\ {bx+ay=-8}\end{cases}​和\begin{cases}{2x+y=-2​}\\ {3x-y=12}\end{cases}​的解,解方程组\begin{cases}{2x+y=-2​}\\ {3x-y=12}\end{cases}​得\begin{cases}{x=2​}\\ {y=-6}\end{cases}​$
$把\begin{cases}{x=2}\\ {y=-6}\end{cases}​代入方程组\begin{cases}{ax+by=-4​}\\ {bx-ay=-8}\end{cases}​得\begin{cases}{2a-6b=-4}\\ {2b-6a=-8}\end{cases}​$
$解得,\begin{cases}{a=\frac{7}{4}}\\ {b=\frac{5}{4}}\end{cases}​$
$∴(5a+b)^2=(5×\frac{7}{4}+\frac{5}{4})^2=10^2=100$
$解:记\begin{cases}{x-y=4a-3①​}\\ {x+2y=-5a②}\end{cases}​$
$由②-①,得2y-(-y)=-5a-(4a-3),$
$所以y=-3a+1$
$把y=-3a+1代入①,得x-(-3a+1)=4a-3$
$所以x=a-2$
$因为x,y互为相反数$
$所以x+y=0$
$即a-2+(-3a+1)=0$
$解得a=-\frac{1}{2}$
$解:记\begin{cases}{2x+y=3m①}\\ {x-4y=-2m②}\end{cases}​$
$由①+②,得3x-3y=m,即x-y=\frac{1}{3}m$
$将y+2m=1+x变形,得x-y=2m-1$
$∴\frac{1}{3}m=2m-1$
$解得m=\frac{3}{5}$