\(\Delta=\left(-2m\right)^2-4.\left(2m^2-1\right)\)
\(=4m^2-8m^2+4\)
\(=4-4m^2\ge0\forall m\)
Theo Vi-ét ta có:
\(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=2m^2-1\end{matrix}\right.\)
Ta có:
\(x^3_1-x^2_1+x^3_2-x^2_2=2\)
\(\Leftrightarrow x^3_1+x^3_2-\left(x^2_1+x^2_2\right)-2=0\)
\(\Leftrightarrow\left(x_1+x_2\right)\left(x^2_1-x_1x_2+x^2_2\right)-\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-2=0\)
\(\Leftrightarrow\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-2x_1x_2-x_1x_2\right]-\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-2=0\)
\(\Leftrightarrow2m\left[\left(2m\right)^2-3\left(2m^2-1\right)\right]-\left[\left(2m^2\right)-2\left(2m^2-1\right)\right]-2=0\)
\(\Leftrightarrow2m\left(4m^2-6m^2+1\right)-4m^2+4m^2-2-2=0\)
\(\Leftrightarrow2m\left(-2m^2+1\right)-4=0\)
\(\Leftrightarrow-4m^3+2m-4=0\)
\(\Leftrightarrow4m^3-2m+4=0\)
\(\Leftrightarrow2\left(2m^2-m\right)=-4\)
\(\Leftrightarrow2m^2-m=-2\)
\(\Leftrightarrow2m^2-m+2=0\)
\(\Delta=\left(-1\right)^2-4.2.2=-15< 0\Rightarrow\) Vô no.
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