Đáp án: $x = \dfrac{7}{2}$
Giải thích các bước giải:
$\begin{array}{l}
a)Dkxd:x\# 2;x\# 3;x\# 4;x\# 5\\
\dfrac{{{x^2} - 4x + 5}}{{x - 2}} + \dfrac{{{x^2} - 10x + 26}}{{x - 5}} = \dfrac{{{x^2} - 6x + 10}}{{x - 3}} + \dfrac{{{x^2} - 8x + 17}}{{x - 4}}\\
\Leftrightarrow \dfrac{{{{\left( {x - 2} \right)}^2} + 1}}{{x - 2}} + \dfrac{{{{\left( {x - 5} \right)}^2} + 1}}{{x - 5}} = \dfrac{{{{\left( {x - 3} \right)}^2} + 1}}{{x - 3}} + \dfrac{{{{\left( {x - 4} \right)}^2} + 1}}{{x - 4}}\\
\Leftrightarrow x - 2 + \dfrac{1}{{x - 2}} + x - 5 + \dfrac{1}{{x - 5}} = x - 3 + \dfrac{1}{{x - 3}} + x - 4 + \dfrac{1}{{x - 4}}\\
\Leftrightarrow \dfrac{1}{{x - 2}} + \dfrac{1}{{x - 5}} = \dfrac{1}{{x - 3}} + \dfrac{1}{{x - 4}}\\
\Leftrightarrow \dfrac{{x - 5 + x - 2}}{{\left( {x - 2} \right)\left( {x - 5} \right)}} - \dfrac{{x - 3 + x - 4}}{{\left( {x - 3} \right)\left( {x - 4} \right)}} = 0\\
\Leftrightarrow \left( {2x - 7} \right)\left( {\dfrac{1}{{\left( {x - 2} \right)\left( {x - 5} \right)}} - \dfrac{1}{{\left( {x - 3} \right)\left( {x - 4} \right)}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{2}\left( {tmdk} \right)\\
\left( {x - 2} \right)\left( {x - 5} \right) - \left( {x - 3} \right)\left( {x - 4} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
{x^2} - 7x + 10 - \left( {{x^2} - 7x + 12} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
- 2 = 0\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{7}{2}
\end{array}$
