Chọn A
Ta có \(x^{2} -xy+3=0\Leftrightarrow y=\frac{x^{2} +3}{x} .\)
Lại có \(2x+3y\le 14\Rightarrow 2x+3\left(\frac{x^{2} +3}{x} \right)\le 14\Leftrightarrow 5x^{2} -14x+9\le 0\Leftrightarrow 1\le x\le \frac{9}{5} .\)
Khi đó
\(P=3x^{2} \left(\frac{x^{2} +3}{x} \right)-x\left(\frac{x^{2} +3}{x} \right)^{2} -2x\left(x^{2} -1\right)
=3x\left(x^{2} +3\right)-x\left(\frac{\left(x^{2} +3\right)^{2} }{x^{2} } \right)-2x^{3} +2x=5x-\frac{9}{x}\) .
Có \(P'=5+\frac{9}{x^{2} } >0\, ,\, \forall x\in \left[1;\frac{9}{5} \right]\Rightarrow P\eqref{GrindEQ__1_}\le P(x)\le P(\frac{9}{5} ),\forall x\in \left[1;\frac{9}{5} \right]\)
Nên \(m=P\left(1\right)=-4;M=P\left(\frac{9}{5} \right)=4\Rightarrow T=2.4-\left(-4\right)=12.\)