Đáp án:
a. x=3
b. x=2017
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne \{ 1;2\} \\
\frac{{2x - 5}}{{x - 2}} - \frac{{3x - 5}}{{x - 1}} = - 1\\
\to (2x - 5)(x - 1) - (3x - 5)(x - 2) = - (x - 2)(x - 1)\\
\to 2{x^2} - 7x + 5 - 3{x^2} + 11x - 10 = - {x^2} + 3x - 2\\
\to x = 3\left( {TM} \right)\\
b.\frac{{x - 3}}{{2014}} - 1 + \frac{{x - 2}}{{2015}} - 1 = \frac{{x - 1}}{{2016}} - 1 + \frac{x}{{2017}} - 1\\
\to \frac{{x - 2017}}{{2014}} + \frac{{x - 2017}}{{2015}} = \frac{{x - 2017}}{{2016}} + \frac{{x - 2017}}{{2017}}\\
\to \left( {x - 2017} \right)\left( {\frac{1}{{2014}} + \frac{1}{{2015}} - \frac{1}{{2016}} - \frac{1}{{2017}}} \right) = 0\\
\to x = 2017
\end{array}\)
