Đáp án:
\(\begin{array}{l}
1)\dfrac{1}{{12y}}\\
2)\dfrac{{2x}}{3}\\
3) - \dfrac{{25}}{{3{x^2}y}}\\
4)\dfrac{5}{{2{x^2} + 14}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne 0;y \ne 0\\
\dfrac{{5{x^2}}}{{42x{y^2}}}.\dfrac{{7y}}{{10x}} = \dfrac{1}{{2.6.y}} = \dfrac{1}{{12y}}\\
2)DK:x \ne \pm y\\
\dfrac{{2{x^2} + 2xy}}{{3\left( {y - x} \right)}}.\dfrac{{y - x}}{{y + x}}\\
= \dfrac{{2x\left( {x + y} \right)}}{{3\left( {y - x} \right)}}.\dfrac{{y - x}}{{y + x}}\\
= \dfrac{{2x}}{3}\\
3)DK:x \ne 0;y \ne 0\\
\dfrac{{20x}}{{3{y^2}}}:\left( { - \dfrac{{4{x^3}}}{{5y}}} \right)\\
= \dfrac{{20x}}{{3{y^2}}}.\left( { - \dfrac{{5y}}{{4{x^3}}}} \right)\\
= - \dfrac{{5.5}}{{3.{x^2}y}} = - \dfrac{{25}}{{3{x^2}y}}\\
4)DK:x \ne 2\\
\dfrac{{5\left( {x - 2} \right)}}{{{x^2} + 7}}.\dfrac{1}{{2\left( {x - 2} \right)}} = \dfrac{5}{{2\left( {{x^2} + 7} \right)}}\\
= \dfrac{5}{{2{x^2} + 14}}
\end{array}\)
