Đáp án:

\(\begin{array}{l}
1)\dfrac{1}{{{x^2} - 2x + 1}} = \dfrac{{5x}}{{5x.{{\left( {x - 1} \right)}^2}}}\\
\dfrac{3}{{5x\left( {x - 1} \right)}} = \dfrac{{3x - 3}}{{5x.{{\left( {x - 1} \right)}^2}}}\\
2)\dfrac{1}{{2\left( {x - 1} \right)}} = \dfrac{{x + 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
\dfrac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{4}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
3)\dfrac{3}{{x - 3}} = \dfrac{{3x + 9}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\
\dfrac{1}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\
4)\dfrac{{5x}}{{2\left( {x + 2} \right)}} = \dfrac{{10{x^2} - 10x}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}\\
\dfrac{{x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{2x + 4}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}\\
5)\dfrac{x}{{2\left( {x + 4} \right)}} = \dfrac{{{x^2} - 4x}}{{2\left( {x + 4} \right)\left( {x - 4} \right)}}\\
\dfrac{{2x + 4}}{{2\left( {x - 4} \right)\left( {x + 4} \right)}}\\
6)\dfrac{{7x}}{{{{\left( {x - 2} \right)}^2}}} = \dfrac{{21{x^2}}}{{3x{{\left( {x - 2} \right)}^2}}}\\
\dfrac{x}{{3x\left( {x - 2} \right)}} = \dfrac{{{x^2} - 2x}}{{3x{{\left( {x - 2} \right)}^2}}}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)\dfrac{1}{{{x^2} - 2x + 1}} = \dfrac{{5x}}{{5x.{{\left( {x - 1} \right)}^2}}}\\
\dfrac{3}{{5x\left( {x - 1} \right)}} = \dfrac{{3\left( {x - 1} \right)}}{{5x.{{\left( {x - 1} \right)}^2}}} = \dfrac{{3x - 3}}{{5x.{{\left( {x - 1} \right)}^2}}}\\
2)\dfrac{1}{{2\left( {x - 1} \right)}} = \dfrac{{x + 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
\dfrac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{4}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
3)\dfrac{3}{{x - 3}} = \dfrac{{3\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{{3x + 9}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\
\dfrac{1}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\
4)\dfrac{{5x}}{{2\left( {x + 2} \right)}} = \dfrac{{5x\left( {x - 2} \right)}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}} = \dfrac{{10{x^2} - 10x}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}\\
\dfrac{{x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{2x + 4}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}\\
5)\dfrac{x}{{2\left( {x + 4} \right)}} = \dfrac{{x\left( {x - 4} \right)}}{{2\left( {x + 4} \right)\left( {x - 4} \right)}} = \dfrac{{{x^2} - 4x}}{{2\left( {x + 4} \right)\left( {x - 4} \right)}}\\
\dfrac{{2x + 4}}{{2\left( {x - 4} \right)\left( {x + 4} \right)}}\\
6)\dfrac{{7x}}{{{{\left( {x - 2} \right)}^2}}} = \dfrac{{21{x^2}}}{{3x{{\left( {x - 2} \right)}^2}}}\\
\dfrac{x}{{3x\left( {x - 2} \right)}} = \dfrac{{x\left( {x - 2} \right)}}{{3x{{\left( {x - 2} \right)}^2}}} = \dfrac{{{x^2} - 2x}}{{3x{{\left( {x - 2} \right)}^2}}}
\end{array}\)