Đáp án:
a. \(x \ge - \dfrac{{17}}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\dfrac{{x - 2}}{6} + \dfrac{{2x + 3}}{4} \ge \dfrac{{x - 3}}{3}\\
\to \dfrac{{2x - 4 + 6x + 9}}{{12}} \ge \dfrac{{4x - 12}}{{12}}\\
\to 8x + 5 \ge 4x - 12\\
\to 4x \ge - 17\\
\to x \ge - \dfrac{{17}}{4}\\
b.DK:x \ne \pm 5\\
\dfrac{x}{{x - 5}} + \dfrac{x}{{x + 5}} = \dfrac{{10x}}{{{x^2} - 25}}\\
\to x\left( {x + 5} \right) + x\left( {x - 5} \right) - 10x = 0\\
\to {x^2} + 5x + {x^2} - 5x - 10x = 0\\
\to 2{x^2} - 10x = 0\\
\to 2x\left( {x - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 5\left( l \right)
\end{array} \right.
\end{array}\)
