\(a^2x-ab=b^2.\left(x-1\right)\)
\(\Leftrightarrow a^2x-ab-b^2.\left(x-1\right)=0\)
\(\Leftrightarrow a^2x-ab-\left(b^2x-b^2\right)=0\)
\(\Leftrightarrow a^2x-ab-b^2x+b^2=0\)
\(\Leftrightarrow\left(a^2x-b^2x\right)-\left(ab-b^2\right)=0\)
\(\Leftrightarrow x.\left(a^2-b^2\right)-b.\left(a-b\right)=0\)
\(\Leftrightarrow x.\left(a-b\right).\left(a+b\right)-b.\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right).\left(x+a+b-b\right)=0\)
\(\Leftrightarrow\left(a-b\right).\left(x+a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\x+a=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0+b\\a=0-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b\\a=-x\end{matrix}\right.\Leftrightarrow a=b=-x.\)
Vậy \(a=b=-x\) thì \(a^2x-ab=b^2.\left(x-1\right).\)
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