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

Đề bài

Thực hiện phép tính:

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

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

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

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

Lời giải chi tiết

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

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

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

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