Ta có \(\Delta=4m^2-4m+1-4m^2+4=-4m+5\)
Để phương trình có hai nghiệm phân biệt thì \(\Delta>0\Rightarrow-4m+5>0\Leftrightarrow m\le\frac{5}{4}\)
Theo hệ thức vi-ét ta có:\(\left\{{}\begin{matrix}a+b=2m-1\\ab=m^2-1\end{matrix}\right.\)
\(\Rightarrow\left(a-b\right)^2=\left(a+b\right)^2-4ab=\left(2m-1\right)^2-4\left(m^2-1\right)=-4m+5\)
\(a-3b=a+b-4b=2m-1-4b\)
\(\left(a-b\right)^2=a-3b\Leftrightarrow-4m+5=2m-1-4b\Leftrightarrow-3m+3=-2b\Leftrightarrow b=\frac{3m-3}{2}\)
\(\Rightarrow a+b=2m-1\Leftrightarrow a+\frac{3m-3}{2}=2m-1\Leftrightarrow a=\frac{m+1}{2}\)
\(\Rightarrow ab=m^2-1\Leftrightarrow\frac{\left(3m-3\right)\left(m+1\right)}{4}=m^2-1\)
\(\Leftrightarrow3m^2-3=4m^2-4\)
\(\Leftrightarrow m^2=1\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-1\end{matrix}\right.\left(tm\right)\)
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