Đáp án:
C2:
b. \(\dfrac{{ - {x^2}y}}{{{x^2} - {y^2}}}\)
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
a.DK:x \ne 4\\
\dfrac{{\left( {4x - 1} \right)\left( {4x + 1} \right)}}{{{{\left( {4x - 1} \right)}^2}}} = \dfrac{{4x + 1}}{{4x - 1}}\\
b.DK:y \ne \pm 2x\\
\dfrac{{{{\left( {2x - y} \right)}^2}}}{{\left( {y - 2x} \right)\left( {y + 2x} \right)}} = \dfrac{{2x - y}}{{ - 2x - y}}\\
C2:\\
a.DK:a \ne 1\\
\dfrac{{3{a^2} - a + 3 + \left( {1 - a} \right)\left( {a - 1} \right) - 2\left( {{a^2} + a + 1} \right)}}{{\left( {a - 1} \right)\left( {{a^2} + a + 1} \right)}}\\
= \dfrac{{3{a^2} - a + 3 - {a^2} + 2a - 1 - 2{a^2} - 2a - 1}}{{\left( {a - 1} \right)\left( {{a^2} + a + 1} \right)}}\\
= \dfrac{{ - a + 1}}{{\left( {a - 1} \right)\left( {{a^2} + a + 1} \right)}}\\
= \dfrac{{ - 1}}{{{a^2} + a + 1}}\\
b.DK:x \ne \pm y\\
\dfrac{{x\left( {{x^2} - {y^2}} \right) - xy\left( {x - y} \right) - {x^3}}}{{\left( {x - y} \right)\left( {x + y} \right)}}\\
= \dfrac{{{x^3} - x{y^2} - {x^2}y + x{y^2} - {x^3}}}{{\left( {x - y} \right)\left( {x + y} \right)}}\\
= \dfrac{{ - {x^2}y}}{{{x^2} - {y^2}}}
\end{array}\)
