Số nghiệm của phương trình \(\dfrac{{2{x^2} - x - 3}}{{\left( {2x - 3} \right)\left( {{x^2} + x + 1} \right)}} - \dfrac{{2{x^2} - 5x + 3}}{{\left( {2x - 3} \right)\left( {{x^2} - x + 1} \right)}} = \dfrac{{6x - 9}}{{x\left( {2x - 3} \right)\left( {{x^4} + {x^2} + 1} \right)}}\) là:

ĐK: \(\left\{ \begin{array}{l}\left( {2x - 3} \right)\left( {{x^2} + x + 1} \right) \ne 0\\\left( {2x - 3} \right)\left( {{x^2} - x + 1} \right) \ne 0\\x\left( {2x - 3} \right)\left( {{x^4} + {x^2} + 1} \right) \ne 0\end{array} \right.\,\left( * \right)\)

Ta có: \({x^2} + x + 1\) \( = {x^2} + 2.\dfrac{1}{2}.x + \dfrac{1}{4} + \dfrac{3}{4}\) \( = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\)

\({x^2} - x + 1\) \( = {x^2} - 2.\dfrac{1}{2}.x + \dfrac{1}{4} + \dfrac{3}{4}\) \( = {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\)

\({x^4} + {x^2} + 1 > 0\)

Do đó \(\left( * \right) \Leftrightarrow \left\{ \begin{array}{l}x \ne \dfrac{3}{2}\\x \ne 0\end{array} \right.\)

Khi đó, \(pt \Leftrightarrow \dfrac{{\left( {2x - 3} \right)\left( {x + 1} \right)}}{{\left( {2x - 3} \right)\left( {{x^2} + x + 1} \right)}} - \dfrac{{\left( {2x - 3} \right)\left( {x - 1} \right)}}{{\left( {2x - 3} \right)\left( {{x^2} - x + 1} \right)}} = \dfrac{{3\left( {2x - 3} \right)}}{{x\left( {2x - 3} \right)\left( {{x^4} + {x^2} + 1} \right)}}\)

\( \Rightarrow \dfrac{{x + 1}}{{{x^2} + x + 1}} - \dfrac{{x - 1}}{{{x^2} - x + 1}} = \dfrac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}}\)

\( \Leftrightarrow \dfrac{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{\left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 1} \right)}} - \dfrac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}}\)

\( \Leftrightarrow \dfrac{{{x^3} + 1}}{{{{\left( {{x^2} + 1} \right)}^2} - {x^2}}} - \dfrac{{{x^3} - 1}}{{{{\left( {{x^2} + 1} \right)}^2} - {x^2}}} = \dfrac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}}\)

\( \Leftrightarrow \dfrac{{{x^3} + 1 - {x^3} + 1}}{{{x^4} + {x^2} + 1}} = \dfrac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}}\)

\( \Leftrightarrow \dfrac{2}{{{x^4} + {x^2} + 1}} = \dfrac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}}\)

\( \Leftrightarrow \dfrac{{2x}}{{x\left( {{x^4} + {x^2} + 1} \right)}} = \dfrac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}}\)

\( \Rightarrow 2x = 3 \Leftrightarrow x = \dfrac{3}{2}\left( {loại} \right)\)

Vậy phương trình đã cho vô nghiệm.