Đáp án:
c) \(x = \dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ {1;2} \right\}\\
1 + \dfrac{{2x - 5}}{{x - 2}} - \dfrac{{3x - 5}}{{x - 1}} = 0\\
\to \dfrac{{\left( {x - 1} \right)\left( {x - 2} \right) + \left( {2x - 5} \right)\left( {x - 1} \right) - \left( {3x - 5} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = 0\\
\to {x^2} - 3x + 2 + 2{x^2} - 7x + 5 - 3{x^2} + 11x - 10 = 0\\
\to x = 3\\
b)DK:x \ne \left\{ {2;4} \right\}\\
\dfrac{{\left( {x - 3} \right)\left( {x - 4} \right) - {{\left( {x - 2} \right)}^2} + \left( {x - 2} \right)\left( {x - 4} \right)}}{{\left( {x - 2} \right)\left( {x - 4} \right)}} = 0\\
\to {x^2} - 7x + 12 - {x^2} + 4x - 4 + {x^2} - 6x + 8 = 0\\
\to {x^2} - 9x + 16 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{9 + \sqrt {17} }}{2}\\
x = \dfrac{{9 - \sqrt {17} }}{2}
\end{array} \right.\\
c)DK:x \ne \left\{ {\dfrac{2}{3};7} \right\}\\
\dfrac{{3x + 2}}{{x - 7}} = \dfrac{{6x + 1}}{{2x - 3}}\\
\to \dfrac{{\left( {3x + 2} \right)\left( {2x - 3} \right) - \left( {6x + 1} \right)\left( {x - 7} \right)}}{{\left( {x - 7} \right)\left( {2x - 3} \right)}} = 0\\
\to 6{x^2} - 5x - 6 - 6{x^2} + x + 7 = 0\\
\to - 4x + 1 = 0\\
\to x = \dfrac{1}{4}\\
d)DK:x \ne \pm 2\\
\dfrac{{\left( {x + 1} \right)\left( {x + 2} \right) - \left( {x - 1} \right)\left( {x - 2} \right) - 2{x^2} - 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} + 3x + 2 - {x^2} + 3x - 2 - 2{x^2} - 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{ - 2{x^2} + 6x - 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}
\end{array}\)
