\(\Rightarrow\left(mx+3\right)^2=\left(2x-m\right)^2\)
\(\Leftrightarrow m^2x^2+6mx+9=4x^2-4mx+m^2\)
\(\Leftrightarrow\left(m^2-4\right)x^2+10mx-m^2+9=0\)
Xét \(m^2-4=0\Leftrightarrow\left[{}\begin{matrix}m=2\\m=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}20x-4+9=0\\-20x-4+9=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=\frac{1}{4}\end{matrix}\right.\)
Xét \(m^2-4\ne0\Leftrightarrow\left\{{}\begin{matrix}m\ne2\\m\ne-2\end{matrix}\right.\)
Để pt có 2 n0 pb<=> \(\Delta'>0\Leftrightarrow25m^2-\left(9-m^2\right)\left(m^2-4\right)>0\)
\(\Leftrightarrow25m^2-9m^2+36+m^4-4m^2>0\)
\(\Leftrightarrow m^4+12m^2+36>0\)
Đặt m2= t
\(\Rightarrow t^2+12t+36>0\)
\(\Leftrightarrow\left(t+6\right)^2>0\left(lđ\right)\)
\(\Rightarrow m^4+12m^2+36>0\)
Vậy vs \(m\ne\pm2\) thì pt luôn có 2 n0 pb
P/s: xem lại hộ tui xem chỗ nèo soai ko nhe =>
