Δ' = (-3)^2 - 2m - 1 = 8 - 2m
pt có 2 nghiệm phân biệt
=>Δ' > 0
=> 8 - 2m > 0 <=> m < 4
viet:
\(\left\{{}\begin{matrix}x1+x2=6\\x1\cdot x2=2m+1\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x1=6-x2\\x1\cdot x2=2m+1\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x1^2=\left(6-x2\right)^2\\x1\cdot x2=2m+1\end{matrix}\right.\)
lại có: x1^2 = x2 - 4
=> \(x2-4=\left(6-x2\right)^2\)
<=> x2 - 4 = 36 - 12x2 + x2^2
<=> \(\left[{}\begin{matrix}x2=5\\x2=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x1=1\\x1=2\end{matrix}\right.\)
mà x1 * x2 = 2m+1
=> m = 2 khi x1 = 1; x2 = 5
m = 7 khi x1 = 2; x2 = 8 (loại)
vậy m = 2 thỏa