Giải phương trình: \(\frac{3}{{x - 3}} + \frac{5}{{x - 5}} + \frac{{17}}{{x - 17}} + \frac{{19}}{{x - 19}} = {x^2} - 11x - 4\).

Điều kiện: x ≠ {3; 5; 17; 19}.

\(\frac{3}{{x - 3}} + \frac{5}{{x - 5}} + \frac{{17}}{{x - 17}} + \frac{{19}}{{x - 19}} = {x^2} - 11x - 4\)

⇔ \(\left( {\frac{x}{{x - 3}} - 1} \right) + \left( {\frac{x}{{x - 5}} - 1} \right) + \left( {\frac{x}{{x - 17}} - 1} \right) + \left( {\frac{x}{{x - 19}} - 1} \right) = x\left( {x - 11} \right)\)– 4

⇔ \(x\left( {\frac{1}{{x - 3}} + \frac{1}{{x - 5}} + \frac{1}{{x - 17}} + \frac{1}{{x - 19}}} \right) = x\left( {x - 11} \right)\)

⇔ \(2x\left( {x - 11} \right)\left( {\frac{1}{{\left( {x - 3} \right)\left( {x - 19} \right)}} + \frac{1}{{\left( {x - 5} \right)\left( {x - 17} \right)}}} \right) = x\left( {x - 11} \right)\)

\( \Leftrightarrow x\left( {x - 11} \right)\left[ {2\left( {\frac{1}{{{x^2} - 22x + 57}} + \frac{1}{{{x^2} - 22x + 85}}} \right) - 1} \right] = 0\)

TH1: x = 0 (t/m)

TH2: x – 11 = 0, suy ra x = 11 (t/m)

TH3: \(\frac{1}{{{x^2} - 22x + 57}} + \frac{1}{{{x^2} - 22x + 85}} = \frac{1}{2}\)

Đặt x² – 22x + 57 = a

Ta có:

\(\frac{1}{a} + \frac{1}{{a + 28}} = \frac{1}{2}\)

⇔ a² + 24a – 56= 0

⇔ \[\left[ \begin{array}{l}a = - 12 + 10\sqrt 2 \\a = - 12 - 10\sqrt 2 \end{array} \right.\]

⇔ \[\left[ \begin{array}{l}{x^2} - 22x + 69 - 10\sqrt 2 = 0\\{x^2} - 22x + 69 + 10\sqrt 2 = 0\end{array} \right.\]

⇔ \[x = 11 \pm \sqrt {52 \pm 10\sqrt 2 } \] (t/m).

Vậy x = 0, x = 11, \[x = 11 \pm \sqrt {52 \pm 10\sqrt 2 } \].