Do \(x=2\) là nghiệm của phương trình nên:

\(\left\{{}\begin{matrix}2a+y=3\\2+ay=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=3-2a\\ay=-3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}ay=3a-2a^2\\ay=-3\end{matrix}\right.\)

\(\Rightarrow3a-2a^2=-3\)

\(\Rightarrow2a^2-3a-3=0\Rightarrow a=\dfrac{3\pm\sqrt{33}}{4}\)

Thay vào ta được 

\(\left\{{}\begin{matrix}a=2a-1\\-1=a^2-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\a^2-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\a=-1\end{matrix}\right.\)

Thay x=1 và y=2 vào HPT, ta được:

\(\left\{{}\begin{matrix}a+2=3\\2+2=a^2+3\end{matrix}\right.\Leftrightarrow a=1\)

Bài 1.

\(\left\{{}\begin{matrix}x-3y=5-2m\\2x+y=3\left(m+1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3y=5-2m\\6x+3y=9m+9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7m+14\\x-3y=5-2m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\m+2-3y=5-2m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\-3y=-3m+3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\y=m-1\end{matrix}\right.\)

\(x_0^2+y_0^2=9m\)

\(\Leftrightarrow\left(m+2\right)^2+\left(m-1\right)^2=9m\)

\(\Leftrightarrow m^2+4m+4+m^2-2m+1-9m=0\)

\(\Leftrightarrow2m^2-7m+5=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}m=1\\m=\dfrac{5}{2}\end{matrix}\right.\) ( Vi-ét )

Với a = 0 ta có \(\left\{{}\begin{matrix}-2y=0\\x+y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=0\end{matrix}\right.\)( không thỏa mãn đề bài )

Với a ≠ 0 ta có : \(\left\{{}\begin{matrix}a^2x-2y=0\\x+y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2\left(4-y\right)-2y=0\\x=4-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a^2-a^2y-2y=0\\x=4-y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a^2+2\right)y=-4a^2\\x=4-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-4a^2}{a^2+2}\\x=4+\dfrac{4a^2}{a^2+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{8a^2+8}{a^2+2}\\y=\dfrac{-4a^2}{a^2+2}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\dfrac{8a^2+8}{a^2+2}\\y=\dfrac{-4a^2}{a^2+2}\end{matrix}\right.\)là nghiệm duy nhất của hệ phương trình 

Để hệ phương trình có nghiệm x = -4 , y = 4a thì :

\(\left\{{}\begin{matrix}\dfrac{8a^2+8}{a^2+2}=-4\\4a=\dfrac{-4a^2}{a^2+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8a^2+8=-4a^2-8\\4a^3+8a=-4a^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a^2+4=0\\a^3+a^2+2a=0\end{matrix}\right.\)( đến đây bạn tự giải nốt rồi kết luận nhé :v ) 

1: Để hệ có nghiệm duy nhất thì \(\dfrac{m}{m-1}\ne\dfrac{1}{-1}\ne-1\)

=>\(\dfrac{m+m-1}{m-1}\ne0\)

=>\(\dfrac{2m-1}{m-1}\ne0\)

=>\(m\notin\left\{\dfrac{1}{2};1\right\}\)(1)

\(\left\{{}\begin{matrix}mx+y=3\\\left(m-1\right)x-y=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}mx+\left(m-1\right)x=3+7\\mx+y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(2m-1\right)=10\\mx+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10}{2m-1}\\y=3-mx=3-\dfrac{10m}{2m-1}=\dfrac{6m-3-10m}{2m-1}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{10}{2m-1}\\y=\dfrac{-4m-3}{2m-1}\end{matrix}\right.\)

Để x và y trái dấu thì x*y<0

=>\(\dfrac{10}{2m-1}\cdot\dfrac{-4m-3}{2m-1}< 0\)

=>\(\dfrac{10\left(4m+3\right)}{\left(2m-1\right)^2}>0\)

=>4m+3>0

=>m>-3/4

Kết hợp (1), ta được: \(\left\{{}\begin{matrix}m>-\dfrac{3}{4}\\m\notin\left\{\dfrac{1}{2};1\right\}\end{matrix}\right.\)

2: Để x,y là số nguyên thì \(\left\{{}\begin{matrix}10⋮2m-1\\-4m-3⋮2m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2m-1\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\\-4m+2-5⋮2m-1\end{matrix}\right.\)

=>\(2m-1\in\left\{1;-1;5;-5\right\}\)

=>\(2m\in\left\{2;0;6;-4\right\}\)

=>\(m\in\left\{1;0;3;-2\right\}\)

Kết hợp (1), ta được: \(m\in\left\{0;3;-2\right\}\)

a, \(\left(I\right):\left\{{}\begin{matrix}2x+ay=b\\ax-by=1\end{matrix}\right.\)

Thay (x;y)=(1;-3) vào hpt có :

\(\left\{{}\begin{matrix}2-3a=b\\a+3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+b=2\\a+3b=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}9a+3b=6\\a+3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8a=5\\a+3b=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5}{8}\\\dfrac{5}{8}+3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5}{8}\\b=\dfrac{1}{8}\end{matrix}\right.\)

Vậy a=5/8 , b=1/8