Mọi người ơi, giúp em bài 3 với ạ. Em cảm ơn nhiều.

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\dfrac{{x + 1}}{{2004}} + \dfrac{{x + 2}}{{2003}} = \dfrac{{x + 3}}{{2002}} + \dfrac{{x + 4}}{{2001}}\\
 \Leftrightarrow \left( {\dfrac{{x + 1}}{{2004}} + 1} \right) + \left( {\dfrac{{x + 2}}{{2003}} + 1} \right) = \left( {\dfrac{{x + 3}}{{2002}} + 1} \right) + \left( {\dfrac{{x + 4}}{{2001}} + 1} \right)\\
 \Leftrightarrow \dfrac{{x + 1 + 2004}}{{2004}} + \dfrac{{x + 2 + 2003}}{{2003}} = \dfrac{{x + 3 + 2002}}{{2002}} + \dfrac{{x + 4 + 2001}}{{2001}}\\
 \Leftrightarrow \dfrac{{x + 2005}}{{2004}} + \dfrac{{x + 2005}}{{2003}} = \dfrac{{x + 2005}}{{2002}} + \dfrac{{x + 2005}}{{2001}}\\
 \Leftrightarrow \left( {x + 2005} \right).\left( {\dfrac{1}{{2004}} + \dfrac{1}{{2003}} - \dfrac{1}{{2002}} - \dfrac{1}{{2001}}} \right) = 0\\
 \Leftrightarrow x + 2005 = 0\\
 \Leftrightarrow x =  - 2005\\
b,\\
\dfrac{{201 - x}}{{99}} + \dfrac{{203 - x}}{{97}} + \dfrac{{205 - x}}{{95}} + 3 = 0\\
 \Leftrightarrow \left( {\dfrac{{201 - x}}{{99}} + 1} \right) + \left( {\dfrac{{203 - x}}{{97}} + 1} \right) + \left( {\dfrac{{205 - x}}{{95}} + 1} \right) = 0\\
 \Leftrightarrow \dfrac{{201 - x + 99}}{{99}} + \dfrac{{203 - x + 97}}{{97}} + \dfrac{{205 - x + 95}}{{95}} = 0\\
 \Leftrightarrow \dfrac{{300 - x}}{{99}} + \dfrac{{300 - x}}{{97}} + \dfrac{{300 - x}}{{95}} = 0\\
 \Leftrightarrow \left( {300 - x} \right)\left( {\dfrac{1}{{99}} + \dfrac{1}{{97}} + \dfrac{1}{{95}}} \right) = 0\\
 \Leftrightarrow 300 - x = 0\\
 \Leftrightarrow x = 300\\
c,\\
\dfrac{{x - 45}}{{55}} + \dfrac{{x - 47}}{{53}} = \dfrac{{x - 55}}{{45}} + \dfrac{{x - 53}}{{47}}\\
 \Leftrightarrow \left( {\dfrac{{x - 45}}{{55}} - 1} \right) + \left( {\dfrac{{x - 47}}{{53}} - 1} \right) = \left( {\dfrac{{x - 55}}{{45}} - 1} \right) + \left( {\dfrac{{x - 53}}{{47}} - 1} \right)\\
 \Leftrightarrow \dfrac{{x - 45 - 55}}{{55}} + \dfrac{{x - 47 - 53}}{{53}} = \dfrac{{x - 55 - 45}}{{45}} + \dfrac{{x - 53 - 47}}{{47}}\\
 \Leftrightarrow \dfrac{{x - 100}}{{55}} + \dfrac{{x - 100}}{{53}} = \dfrac{{x - 100}}{{45}} + \dfrac{{x - 100}}{{47}}\\
 \Leftrightarrow \left( {x - 100} \right)\left( {\dfrac{1}{{55}} + \dfrac{1}{{53}} - \dfrac{1}{{45}} - \dfrac{1}{{47}}} \right) = 0\\
 \Leftrightarrow x - 100 = 0\\
 \Leftrightarrow x = 100\\
d,\\
\dfrac{{2x + 1}}{{2x - 1}} - \dfrac{{2x - 1}}{{2x + 1}} = \dfrac{8}{{4{x^2} - 1}}\,\,\,\,\,\left( {x \ne  \pm \dfrac{1}{2}} \right)\\
 \Leftrightarrow \dfrac{{{{\left( {2x + 1} \right)}^2} - {{\left( {2x - 1} \right)}^2}}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}} = \dfrac{8}{{4{x^2} - 1}}\\
 \Leftrightarrow \dfrac{{\left( {4{x^2} + 4x + 1} \right) - \left( {4{x^2} - 4x + 1} \right)}}{{4{x^2} - 1}} = \dfrac{8}{{4{x^2} - 1}}\\
 \Leftrightarrow \dfrac{{8x}}{{4{x^2} - 1}} = \dfrac{8}{{4{x^2} - 1}}\\
 \Leftrightarrow 8x = 8\\
 \Leftrightarrow x = 1\\
e,\\
\dfrac{6}{{{x^2} - 1}} + 5 = \dfrac{{8x - 1}}{{4x + 4}} - \dfrac{{12x - 1}}{{4 - 4x}}\,\,\,\,\,\left( {x \ne  \pm 1} \right)\\
 \Leftrightarrow \dfrac{6}{{{x^2} - 1}} + 5 = \dfrac{{8x - 1}}{{4\left( {x + 1} \right)}} + \dfrac{{12x - 1}}{{4.\left( {x - 1} \right)}}\\
 \Leftrightarrow \dfrac{{6 + 5.\left( {{x^2} - 1} \right)}}{{{x^2} - 1}} = \dfrac{{\left( {8x - 1} \right)\left( {x - 1} \right) + \left( {12x - 1} \right)\left( {x + 1} \right)}}{{4\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 \Leftrightarrow \dfrac{{5{x^2} + 1}}{{{x^2} - 1}} = \dfrac{{\left( {8{x^2} - 9x + 1} \right) + \left( {12{x^2} + 11x - 1} \right)}}{{4\left( {{x^2} - 1} \right)}}\\
 \Leftrightarrow \dfrac{{5{x^2} + 1}}{{{x^2} - 1}} = \dfrac{{20{x^2} + 2x}}{{4\left( {{x^2} - 1} \right)}}\\
 \Leftrightarrow 4\left( {5{x^2} + 1} \right) = 20{x^2} + 2x\\
 \Leftrightarrow 4 = 2x\\
 \Leftrightarrow x = 2
\end{array}\)