Đáp án:
\(\begin{array}{l}
x \ge 2 \Rightarrow N = \dfrac{x}{{x + 10}} + 12x - 3\\
x < 2 \Rightarrow N = \dfrac{{ - x}}{{x + 10}} + 12x - 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}N = \dfrac{{x\left| {x - 2} \right|}}{{{x^2} + 8x - 20}} + 12x - 3\\N = \dfrac{{x\left| {x - 2} \right|}}{{\left( {x - 2} \right)\left( {x + 10} \right)}} + 12x - 3\end{array}\)
TH1: \(x \ge 2 \Rightarrow \left| {x - 2} \right| = x - 2\)
\(\begin{array}{l} \Rightarrow N = \dfrac{{x\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 10} \right)}} + 12x - 3\\\,\,\,\,\,\,N = \dfrac{x}{{x + 10}} + 12x - 3\end{array}\)
TH2: \(x < 2 \Rightarrow \left| {x - 2} \right| = - \left( {x - 2} \right)\)
\(\begin{array}{l} \Rightarrow N = \dfrac{{ - x\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 10} \right)}} + 12x - 3\\\,\,\,\,\,N = \dfrac{{ - x}}{{x + 10}} + 12x - 3\end{array}\)
