\(\Delta=\left(4m+1\right)^2-4\left(2m-1\right)=16m^2+8m+1-8m+4=16m^2+5>0\)
Vậy pt luôn có 2 nghiệm pb
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=4m+1\\x_1x_2=2m-1\end{matrix}\right.\)
Ta có : \(3x_1^2+3x_2^2-x_1-x_2-48m^2=15\)
\(\Leftrightarrow3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-\left(x_1+x_2\right)-48m^2=15\)
\(\Leftrightarrow3\left(4m+1\right)^2-6\left(2m-1\right)-4m-1-48m^2=15\)
\(\Leftrightarrow48m^2+24m+3-12m+6-4m-1-48m^2=15\)
\(\Leftrightarrow8m=7\Leftrightarrow m=\dfrac{7}{8}\)
\(\Delta=\left[-\left(4m+1\right)^2\right]-4\left(2m-1\right)\\ =16m^2+8m+1-8m+4\\ =16m^2+5>0\)
Suy ra pt luôn có 2 nghiệm phân biệt
Áp dụng định lý Vi-ét ta có:\(\left\{{}\begin{matrix}x_1+x_2=4m+1\\x_1x_2=2m-1\end{matrix}\right.\)
\(3x^2_1+3x^2_2-x_1-x_2-48m^2=15\\ \Leftrightarrow3\left(x^2_1+x_2^2\right)-\left(x_1+x_2\right)-48m^2=15\\ \Leftrightarrow3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-\left(4m+1\right)-48m^2=15\\ \Leftrightarrow3\left(4m+1\right)^2-6\left(2m-1\right)-4m-1-48m^2=15\)
\(\Leftrightarrow3\left(16m^2+8m+1\right)-12m+6-4m-1-48m^2-15=0\)
\(\Leftrightarrow48m^2+24m+3-12m+6-4m-1-48m^2-15=0\)
\(\Leftrightarrow8m-7=0\)
\(\Leftrightarrow m=\dfrac{7}{8}\)
