Bài 2 trang 48 SGK Toán 8 Tập 1 Cánh diều - CD

a) Ta có: \(\dfrac{{20{\rm{x}}}}{{3{y^2}}}:\left( { - \dfrac{{15{{\rm{x}}^2}}}{{6y}}} \right) = \dfrac{{20{\rm{x}}}}{{3{y^2}}}.\left( { - \dfrac{{6y}}{{15{{\rm{x}}^2}}}} \right) = \dfrac{{20{\rm{x}}.\left( { - 6y} \right)}}{{3{y^2}.15{{\rm{x}}^2}}} = \dfrac{{ - 8}}{{3{\rm{x}}y}}\)

b) Ta có: \(\dfrac{{9{{\rm{x}}^2} - {y^2}}}{{x + y}}:\dfrac{{3{\rm{x}} + y}}{{2{\rm{x}} + 2y}} \\= \dfrac{{\left( {3{\rm{x}} - y} \right)\left( {3{\rm{x}} + y} \right)}}{{x + y}}.\dfrac{{2{\rm{x}} + 2y}}{{3{\rm{x}} + y}} \\= \dfrac{{\left( {3{\rm{x}} - y} \right)\left( {3{\rm{x}} + y} \right).2.\left( {x + y} \right)}}{{(x + y).\left( {3{\rm{x}} + y} \right)}} \\= 2\left( {3{\rm{x}} - y} \right)\)

c) Ta có:

\(\begin{array}{l}\dfrac{{{x^3} + {y^3}}}{{y - x}}:\dfrac{{{x^2} - xy + {y^2}}}{{{x^2} - 2{\rm{x}}y + {y^2}}} \\= \dfrac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{y - x}}.\dfrac{{{x^2} - 2{\rm{x}}y + {y^2}}}{{{x^2} - xy + {y^2}}}\\ = \dfrac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right).{{\left( {x - y} \right)}^2}}}{{ - (x - y)\left( {{x^2} - xy + {y^2}} \right)}} \\= - \left( {x + y} \right)\left( {y - x} \right) = - \left( {{y^2} - {x^2}} \right) \\= {x^2} - {y^2}\end{array}\)

d) Ta có: \(\dfrac{{9 - {x^2}}}{x}:\left( {x - 3} \right) \\= \dfrac{{\left( {3 - x} \right)\left( {3 + x} \right)}}{x}.\dfrac{1}{{x - 3}} \\= \dfrac{{ - \left( {x - 3} \right)\left( {3 + x} \right)}}{{x.\left( {x - 3} \right)}} \\= \dfrac{{ - \left( {3 + x} \right)}}{x}.\)

-- Mod Toán 8 HỌC247