Đáp án:
h) \(\dfrac{{2\left( {x - y} \right)}}{{{x^2} + xy + {y^2}}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{2\left( {x + 2} \right)}}{{10}} + \dfrac{{2 - x}}{{15}}\\
= \dfrac{{x + 2}}{5} + \dfrac{{2 - x}}{{15}}\\
= \dfrac{{3x + 6 + 2 - x}}{{15}}\\
= \dfrac{{2x + 8}}{{15}}\\
b)\dfrac{{18x + 4\left( {2x - 1} \right) + 3\left( {2 - x} \right)}}{{60}}\\
= \dfrac{{18x + 8x - 4 + 6 - 3x}}{{60}}\\
= \dfrac{{23x + 2}}{{60}}\\
c)\dfrac{{{x^2} - 5xy}}{{xy}} + \dfrac{{5y - x}}{y} + \dfrac{{x + 2y}}{x}\\
= \dfrac{{{x^2} - 5xy + 5xy - {x^2} + xy + 2{y^2}}}{{xy}}\\
= \dfrac{{xy + 2{y^2}}}{{xy}}\\
= \dfrac{{x + 2y}}{x}\\
d)\dfrac{{x\left( {xy - {x^2}} \right) + \left( {2x - y} \right)\left( {xy - {y^2}} \right)}}{{\left( {xy - {y^2}} \right)\left( {xy - {x^2}} \right)}}\\
= \dfrac{{{x^2}y - {x^3} + 2{x^2}y - 2x{y^2} - x{y^2} + {y^3}}}{{\left( {xy - {y^2}} \right)\left( {xy - {x^2}} \right)}}\\
e)\dfrac{{{{\left( {1 - 2x} \right)}^2} - 4{x^2} + 1}}{{2x\left( {1 - 2x} \right)}}\\
= \dfrac{{1 - 4x + 4{x^2} - 4{x^2} + 1}}{{2x\left( {1 - 2x} \right)}}\\
= \dfrac{{2 - 4x}}{{2x\left( {1 - 2x} \right)}}\\
= \dfrac{1}{x}\\
f)\dfrac{{{x^2} - 2\left( {x + 2} \right) + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} - 2x - 4 + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} - x - 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{\left( {x - 3} \right)\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{x - 3}}{{x - 2}}\\
g)\dfrac{{2x}}{{x\left( {x + 2y} \right)}} + \dfrac{y}{{y\left( {x - 2y} \right)}} + \dfrac{4}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}}\\
= \dfrac{{2\left( {x - 2y} \right) + x + 2y + 4}}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}}\\
= \dfrac{{2x - 4y + x + 2y + 4}}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}}\\
= \dfrac{{3x - 2y + 4}}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}}\\
h)\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} - 2xy + {y^2} + {x^2} - 2xy + {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}\)
