Đáp án:
c. x=6
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne 3\\
\left| {x - 2} \right| = 1\\
\to \left[ \begin{array}{l}
x - 2 = 1\\
x - 2 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\left( l \right)\\
x = 1
\end{array} \right.\\
\to A = \dfrac{1}{{1 - 3}} = - \dfrac{1}{2}\\
b.DK:x \ne \pm 5\\
B = \dfrac{{2x\left( {x - 5} \right) - {x^2} + 15x}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
= \dfrac{{2{x^2} - 10x - {x^2} + 15x}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
= \dfrac{{{x^2} + 5x}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
= \dfrac{x}{{x - 5}}\\
Q = B:A = \dfrac{x}{{x - 5}}:\dfrac{x}{{x - 3}}\\
= \dfrac{x}{{x - 5}}.\dfrac{{x - 3}}{x}\\
= \dfrac{{x - 3}}{{x - 5}}\\
c.Q = 3\\
\to \dfrac{{x - 3}}{{x - 5}} = 3\\
\to x - 3 = 3x - 15\\
\to 2x = 12\\
\to x = 6\left( {TM} \right)
\end{array}\)
