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

Ta có:

\(\begin{array}{l} {x^2} + x + 1 = {x^2} + 2.\frac{1}{2}x + {\left( {\frac{1}{2}} \right)^2} + \frac{3}{4} = {\left( {x + \frac{1}{2}} \right)^2} + \frac{3}{4} > 0\\ {x^2} - x + 1 = {x^2} - 2.\frac{1}{2}x + {\left( {\frac{1}{2}} \right)^2} + \frac{3}{4} = {\left( {x - \frac{1}{2}} \right)^2} + \frac{3}{4} > 0\\ {x^4} + {x^2} + 1 > 0 \end{array}\)

ĐK 

\(\begin{array}{l} \left\{ \begin{array}{l} (2x - 3)({x^2} + x + 1) \ne 0\\ (2x - 3)({x^2} - x + 1) \ne 0\\ x(2x - 3)({x^4} + {x^2} + 1) \ne 0 \end{array} \right. \to \left\{ \begin{array}{l} 2x - 3 \ne 0\\ 2x - 3 \ne 0\\ x(2x - 3) \ne 0 \end{array} \right. \to \left\{ \begin{array}{l} x \ne \frac{3}{2}\\ x \ne \frac{3}{2}\\ \left[ \begin{array}{l} x \ne \frac{3}{2}\\ x \ne 0 \end{array} \right. \end{array} \right. \to \left\{ \begin{array}{l} x \ne \frac{3}{2}\\ x \ne 0 \end{array} \right. \end{array}\)

Do đó

\(\begin{array}{l} pt \Leftrightarrow \frac{{\left( {2x - 3} \right)\left( {x + 1} \right)}}{{\left( {2x - 3} \right)\left( {{x^2} + x + 1} \right)}} - \frac{{\left( {2x - 3} \right)\left( {x - 1} \right)}}{{\left( {2x - 3} \right)\left( {{x^2} - x + 1} \right)}} = \frac{{3\left( {2x - 3} \right)}}{{x\left( {2x - 3} \right)\left( {{x^4} + {x^2} + 1} \right)}}\\ \Rightarrow \frac{{x + 1}}{{{x^2} + x + 1}} - \frac{{x - 1}}{{{x^2} - x + 1}} = \frac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}}\\ \Leftrightarrow \frac{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{\left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 1} \right)}} - \frac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}}\\ \Leftrightarrow \frac{{{x^3} + 1}}{{{{\left( {{x^2} + 1} \right)}^2} - {x^2}}} - \frac{{{x^3} - 1}}{{{{\left( {{x^2} + 1} \right)}^2} - {x^2}}} = \frac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}}\\ \Leftrightarrow \frac{{{x^3} + 1 - {x^3} + 1}}{{{x^4} + {x^2} + 1}} = \frac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}}\\ \Leftrightarrow \frac{2}{{{x^4} + {x^2} + 1}} = \frac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}}\\ \Leftrightarrow \frac{{2x}}{{x\left( {{x^4} + {x^2} + 1} \right)}} = \frac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}}\\ \Rightarrow 2x = 3 \Leftrightarrow x = \frac{3}{2}\left( l \right) \end{array}\)

Vậy phương trình đã cho vô nghiệm.