Đáp án:
\(h)8{y^2}x\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{2x - 2y}}{{x - y}} = \dfrac{{2\left( {x - y} \right)}}{{x - y}} = 2\\
b)\dfrac{{ - 1 + 2x}}{{x - 1}}\\
c)\dfrac{{x + 15 + 2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{x + 15 + 2x - 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{3x + 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{3}{{x - 3}}\\
d)\dfrac{3}{{2\left( {x + 3} \right)}} + \dfrac{{6 - x}}{{2x\left( {x + 3} \right)}}\\
= \dfrac{{3x + 6 - x}}{{2x\left( {x + 3} \right)}} = \dfrac{{2x + 6}}{{2x\left( {x + 3} \right)}} = \dfrac{1}{x}\\
e)\dfrac{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}}{{x\left( {x + 4} \right)}}.\dfrac{{3x}}{{2\left( {1 - 3x} \right)}}\\
= \dfrac{{3\left( {1 + 3x} \right)}}{{2\left( {x + 4} \right)}}\\
f)\dfrac{{x - y}}{{x + y}}.\dfrac{{3\left( {x - y} \right)\left( {x + y} \right)}}{{x\left( {x - y} \right)}}\\
= \dfrac{{3\left( {x - y} \right)}}{x}\\
g)\dfrac{{\left( {x + 3} \right)\left( {x - 3} \right) - {x^2} + 9}}{{x\left( {x - 3} \right)}}\\
= \dfrac{{{x^2} - 9 - {x^2} + 9}}{{x\left( {x - 3} \right)}} = 0\\
h)\dfrac{{4{y^3}}}{{7{x^2}}}.\dfrac{{14{x^3}}}{y} = 8{y^2}x
\end{array}\)
