a) ĐKXĐ: \(x \ne 1\)
\(\begin{array}{*{20}{l}}{\dfrac{{2{\rm{x}} - 1}}{{x - 1}} + 1 = \dfrac{1}{{x - 1}}}\\\begin{array}{l} \Leftrightarrow \dfrac{{2{\rm{x}} - 1}}{{x - 1}} + \dfrac{{x - 1}}{{x - 1}} = \dfrac{1}{{x - 1}}\\ \Rightarrow 2x - 1 + x - 1 = 1\end{array}\\\begin{array}{l} \Leftrightarrow 3{\rm{x}} - 2 = 1\\ \Leftrightarrow 3x = 1 + 2\end{array}\\{ \Leftrightarrow 3{\rm{x}} = 3}\\{ \Leftrightarrow {\rm{x}}{\kern 1pt} {\rm{ = }}{\kern 1pt} {\rm{3:3}}}\\{ \Leftrightarrow {\rm{x}}{\kern 1pt} {\rm{ = }}{\kern 1pt} 1\left( \text{loại} \right)}\end{array}\)
Vậy phương trình vô nghiệm.
b) ĐKXĐ: \(x \ne -1\)
\(\matrix{\dfrac{{5{\text{x}}}}{{2{\text{x}} + 2}} + 1 = - \dfrac{6}{{x + 1}} \hfill \cr { \Leftrightarrow \dfrac{{5{\text{x}}}}{{2\left( {{\text{x}} + 1} \right)}} + 1 = - \dfrac{6}{{x + 1}}} \hfill \cr \matrix{ \Leftrightarrow \dfrac{{5{\text{x}}}}{{2\left( {{\text{x}} + 1} \right)}} + \dfrac{{2x + 2}}{{2\left( {x + 1} \right)}} = - \dfrac{{6.2}}{{2\left( {x + 1} \right)}} \hfill \cr \Rightarrow 5x + 2x + 2 = - 12 \hfill \cr} \hfill \cr { \Leftrightarrow 7{\rm{x}} + 2 = - 12} \hfill \cr { \Leftrightarrow 7{\rm{x}} = - 12 - 2} \hfill \cr { \Leftrightarrow 7{\rm{x}} = - 14} \hfill \cr { \Leftrightarrow x = \left( { - 14} \right):7} \hfill \cr { \Leftrightarrow {\rm{x}}{\kern 1pt} {\rm{ = }} - 2\left( \text{thỏa mãn} \right)} \hfill \cr } \)
Vậy phương trình có nghiệm \(x = -2\).
c) ĐKXĐ: \(x \ne 0\).
\(\begin{array}{l}x + \dfrac{1}{x} = {x^2} + \dfrac{1}{{{x^2}}}\\ \Leftrightarrow \dfrac{{{x^3}}}{{{x^2}}} + \dfrac{x}{{{x^2}}} = \dfrac{{{x^4}}}{{{x^2}}} + \dfrac{1}{{{x^2}}}\\ \Rightarrow {x^3} + x = {x^4} + 1\\ \Leftrightarrow {x^4} - {x^3} - x + 1 = 0\\ \Leftrightarrow {x^3}\left( {x - 1} \right) - \left( {x - 1} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {{x^3} - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x - 1 = 0\\{x^3} - 1 = 0\end{array} \right. \\\Leftrightarrow x = 1\left( \text{thỏa mãn} \right)\end{array}\)
Vậy phương trình có nghiệm duy nhất \(x = 1\).
d) ĐKXĐ: \(x \ne 0; x\ne-1\).
\(\begin{array}{l}\dfrac{{x + 3}}{{x + 1}} + \dfrac{{x - 2}}{x} = 2\\ \Leftrightarrow \dfrac{{x\left( {x + 3} \right)}}{{x\left( {x + 1} \right)}} + \dfrac{{\left( {x - 2} \right)\left( {x + 1} \right)}}{{x\left( {x + 1} \right)}} = \dfrac{{2x\left( {x + 1} \right)}}{{x\left( {x + 1} \right)}} \\\Rightarrow x\left( {x + 3} \right) + \left( {x - 2} \right)\left( {x + 1} \right) = 2x\left( {x + 1} \right)\\\Leftrightarrow {x^2} + 3{\rm{x}} + {x^2} - 2{\rm{x}} + x - 2 = 2{{\rm{x}}^2} + 2{\rm{x}}\\ \Leftrightarrow 2{{\rm{x}}^2} + 2{\rm{x}} - 2\, - 2{{\rm{x}}^2} - 2{\rm{x}} = 0\\ \Leftrightarrow 0x = 2\left( \text{Vô nghiệm} \right)\end{array}\)
Vậy phương trình đã cho vô nghiệm
