Đáp án:
B1:
3) \(\dfrac{{2\left( {x - y} \right)}}{{{x^2} + xy + {y^2}}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
1)DK:x \ne 2y\\
\dfrac{{2x + 1 - 5y - x + y + 1}}{{x - 2y}}\\
= \dfrac{{x - 4y + 2}}{{x - 2y}}\\
2)DK:x \ne \pm 1\\
\dfrac{{{x^2}}}{{x + 1}} + \dfrac{{2x}}{{{x^2} - 1}} - \dfrac{1}{{x - 1}}\\
= \dfrac{{{x^2}\left( {x - 1} \right) + 2x - x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{{x^3} - {x^2} + x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{{x^2}\left( {x - 1} \right) + \left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} + 1}}{{x + 1}}\\
3)DK:x \ne y\\
\dfrac{{{x^2} + xy + {y^2} - 3xy + {{\left( {x - y} \right)}^2}}}{{\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}}\\
= \dfrac{{{x^2} + xy + {y^2} - 3xy + {x^2} - 2xy + {y^2}}}{{\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}}\\
= \dfrac{{2{x^2} - 4xy + 2{y^2}}}{{\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}}\\
= \dfrac{{2{{\left( {x - y} \right)}^2}}}{{\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}}\\
= \dfrac{{2\left( {x - y} \right)}}{{{x^2} + xy + {y^2}}}
\end{array}\)
