Giải bài 3 trang 25 sách bài tập toán 8 – Chân trời sáng tạo

Đề bài

Tính:

a) \(\frac{{{x^2} – 2xy}}{y}.\frac{{{y^2}}}{x}\);                                

b) \(\frac{{{x^2} – 9{y^2}}}{{3x{y^2}}}.\frac{{xy}}{{x + 3y}}\);                           

c) \(\frac{{1 – {x^2}}}{{2x + 4y}}.\frac{{{x^2} + 4xy + 4{y^2}}}{{3 – 3x}}\);         

d) \(\frac{{{x^3} – {y^3}}}{{x + y}}.\frac{{{x^2} – {y^2}}}{{{x^2} + xy + {y^2}}}\)

Lời giải chi tiết

a) \(\frac{{{x^2} – 2xy}}{y}.\frac{{{y^2}}}{x} = \frac{{x\left( {x – 2y} \right).{y^2}}}{{xy}} = y\left( {x – 2y} \right)\);                               

b) \(\frac{{{x^2} – 9{y^2}}}{{3x{y^2}}}.\frac{{xy}}{{x + 3y}} = \frac{{\left( {x – 3y} \right)\left( {x + 3y} \right)xy}}{{3x{y^2}\left( {x + 3y} \right)}} = \frac{{x – 3y}}{{3y}}\);                   

c) \(\frac{{1 – {x^2}}}{{2x + 4y}}.\frac{{{x^2} + 4xy + 4{y^2}}}{{3 – 3x}} = \frac{{\left( {1 – x} \right)\left( {1 + x} \right){{\left( {x + 2y} \right)}^2}}}{{2\left( {x + 2y} \right)3\left( {1 – x} \right)}} = \frac{{\left( {1 + x} \right)\left( {x + 2y} \right)}}{6}\);          

d) \(\frac{{{x^3} – {y^3}}}{{x + y}}.\frac{{{x^2} – {y^2}}}{{{x^2} + xy + {y^2}}} = \frac{{\left( {x – y} \right)\left( {{x^2} + xy + {y^2}} \right)\left( {x – y} \right)\left( {x + y} \right)}}{{\left( {x + y} \right)\left( {{x^2} + xy + {y^2}} \right)}} = {\left( {x – y} \right)^2}\).