a/\(a^2+a^2x=a+x\)
\(\Leftrightarrow a\left(a-1\right)+x\left(a-1\right)\left(a+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(a+x\left(a+1\right)\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+x\left(a+1\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=1\left(1\right)\\x=-\frac{a}{a+1}\left(2\right)\end{matrix}\right.\)
Với (1) x vô số nghiệm
Vậy để x duy nhất thì \(a\ne+-1\)
b/\(\Leftrightarrow2ax-2x-ax+a-2a-3=0\)
\(\Leftrightarrow ax-2x-a+2=5\)
\(\Leftrightarrow x\left(a-2\right)-\left(a-2\right)=5\)
\(\Leftrightarrow\left(x-1\right)\left(a-2\right)=5\)(1)
Với x=1 PT vô nghiệm
\(\left(1\right)\Rightarrow a=\frac{5}{x-1}+2=\frac{2x+3}{x-1}\)
