Giải bài 2 trang 48 SGK Toán 8 tập 1 - Cánh diều

Đề bài

Thực hiện phép tính:

\(a)\dfrac{{20{{x}}}}{{3{y^2}}}:\left( { - \dfrac{{15{{{x}}^2}}}{{6y}}} \right)\)

\(b)\dfrac{{9{{{x}}^2} - {y^2}}}{{x + y}}:\dfrac{{3{{x}} + y}}{{2{{x}} + 2y}}\)

\(c)\dfrac{{{x^3} + {y^3}}}{{y - x}}:\dfrac{{{x^2} - xy + {y^2}}}{{{x^2} - 2{{x}}y + {y^2}}}\)

\(d)\dfrac{{9 - {x^2}}}{x}:\left( {x - 3} \right)\)

Lời giải chi tiết

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

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

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

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