Đáp án:
e) \(\dfrac{{2x}}{{x + 2y}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 2\\
\dfrac{{4\left( {x - 2} \right) + 2\left( {x + 2} \right) - 5x + 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{4x - 8 + 2x + 4 - 5x + 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{1}{{x - 2}}\\
b)DK:x \ne \left\{ {0;\dfrac{1}{2}} \right\}\\
\dfrac{{\left( {1 - 3x} \right)\left( {2x - 1} \right) + \left( {3x - 2} \right).2x - 3x + 2}}{{2x\left( {2x - 1} \right)}}\\
= \dfrac{{ - 6{x^2} + 5x - 1 + 6{x^2} - 4x - 3x + 2}}{{2x\left( {2x - 1} \right)}}\\
= \dfrac{{ - 2x + 1}}{{2x\left( {2x - 1} \right)}} = - \dfrac{1}{{2x}}\\
c)DK:x \ne \pm 3\\
\dfrac{1}{{{{\left( {x + 3} \right)}^2}}} - \dfrac{1}{{{{\left( {x - 3} \right)}^2}}} + \dfrac{x}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} - 6x + 9 - {x^2} - 6x - 9 + {x^3} - 9x}}{{{{\left( {x + 3} \right)}^2}{{\left( {x - 3} \right)}^2}}}\\
= \dfrac{{{x^3} - 21x}}{{{{\left( {x + 3} \right)}^2}{{\left( {x - 3} \right)}^2}}}\\
d)DK:x \ne 1\\
\dfrac{{{x^2} + 2 + 2\left( {x - 1} \right) - {x^2} - x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{1}{{{x^2} + x + 1}}\\
e)DK:x \ne \pm 2y\\
\dfrac{{x\left( {x + 2y} \right) + x\left( {x - 2y} \right) - 4xy}}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}}\\
= \dfrac{{{x^2} + 2xy + {x^2} - 2xy - 4xy}}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}}\\
= \dfrac{{2{x^2} - 4xy}}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}}\\
= \dfrac{{2x}}{{x + 2y}}
\end{array}\)
