Ta có : \(x^2-2\left(m-1\right)x+m^2-2m-3=0\)
=> \(\Delta^,=b^{,2}-ac=\left(m-1\right)^2-\left(m^2-2m-3\right)\)
=> \(\Delta^,=m^2-2m+1-m^2+2m+3=4>0\)
Nên phương trình có 2 nghiệm phân biệt .
\(\left\{{}\begin{matrix}x_1=\frac{-b^,-\sqrt{\Delta^,}}{a}=\frac{m-1-2}{1}=m-3\\x_2=m+1\end{matrix}\right.\)
Ta có : \(\sqrt{x_1}=x_2+4\)
=> \(x^2_2+8x_2+16=x_1\)
TH1 : \(\left\{{}\begin{matrix}x_1=m-3\\x_2=m+1\end{matrix}\right.\)
=> \(\left(m-1\right)^2+8\left(m-1\right)+16=m-3+4=m-3\)
=> \(m^2-2m+1+8m-8+16-m+3=0\)
=> \(m^2+5m+12=0\)
=> \(\left(m^2+\frac{2m.5}{2}+\frac{25}{4}\right)+\frac{23}{4}=\left(m+\frac{5}{2}\right)^2+\frac{23}{4}>0\)
=> PT vô nghiệm .
TH2 : \(\left\{{}\begin{matrix}x_1=m+1\\x_2=m-3\end{matrix}\right.\)
=> \(\left(m-3\right)^2+8\left(m-3\right)+16=m+1\)
=> \(m^2-6m+9+8m-24+16-m-1=0\)
=> \(m^2+m=0\)
=> \(\left[{}\begin{matrix}m=-1\\m=0\end{matrix}\right.\) ( TM )
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