Đáp án:
\[A = \frac{{{x^2}}}{{x - 2}}\,\,\,\,\,\,\,\left( {x \ne 0;\,\,\,x \ne \pm 2} \right)\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ne 0\\
x \ne 2
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
A = \frac{{{x^2} + 2x}}{{{x^2} - 4x + 4}}:\left( {\frac{{x + 2}}{x} - \frac{1}{{2 - x}} + \frac{{6 - {x^2}}}{{{x^2} - 2x}}} \right)\\
= \frac{{x\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}:\left( {\frac{{x + 2}}{x} + \frac{1}{{x - 2}} + \frac{{6 - {x^2}}}{{x\left( {x - 2} \right)}}} \right)\\
= \frac{{x\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}:\frac{{\left( {x + 2} \right)\left( {x - 2} \right) + x + 6 - {x^2}}}{{x\left( {x - 2} \right)}}\\
= \frac{{x\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}:\frac{{{x^2} - 4 + x + 6 - {x^2}}}{{x\left( {x - 2} \right)}}\\
= \frac{{x\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}:\frac{{x + 2}}{{x\left( {x - 2} \right)}}\,\,\,\,\,\,\,\,\,\,\,\left( {x \ne - 2} \right)\\
= \frac{{x\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}.\frac{{x\left( {x - 2} \right)}}{{x + 2}}\\
= \frac{{{x^2}}}{{x - 2}}\\
\Rightarrow A = \frac{{{x^2}}}{{x - 2}}\,\,\,\,\,\,\,\left( {x \ne 0;\,\,\,x \ne \pm 2} \right)
\end{array}\)
