a. Với m = 2 ; p/t : \(x^2-2x-5=0\Leftrightarrow\) \(\left(x-1\right)^2=6\Leftrightarrow\) \(\left[{}\begin{matrix}x=1+\sqrt{6}\\x=1-\sqrt{6}\end{matrix}\right.\)
b. \(\Delta'=\left(m-1\right)^2-\left(-2m-1\right)=m^2-2m+1+2m+1=m^2+2\ge2>0\forall m\)
=> P/t có 2 no p/b x1 ; x2
Theo viet ta có : \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-2m-1\end{matrix}\right.\)
Ta có : \(2x_1+3x_2+3x_1x_2=-11\) \(\Leftrightarrow2\left(x_1+x_2\right)+x_2+3x_1x_2+11=0\)
\(\Leftrightarrow4\left(m-1\right)+x_2-6m-3+11=0\) \(\Leftrightarrow x_2=2m-4\)
Suy ra : \(x_1=-2\)
Khi đó : \(\left(2m-4\right).\left(-2\right)=-2m-1\) \(\Leftrightarrow-4m+8=-2m-1\)
\(\Leftrightarrow m=\dfrac{9}{2}\)
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