Giải bài 5 trang 25 sách bài tập toán 8 – Chân trời sáng tạo

Đề bài

Tính:

a) \(\frac{{{x^2} – 5x}}{{4{y^2}}}:\frac{{5x}}{{2y}}\);                                          

b) \(\frac{{{x^2} – 1}}{y}:\frac{{x + 1}}{{{y^2}}}\);

c) \(\left( {{x^2} – 2xy} \right):\frac{{5x – 10y}}{x}\);                                  

d) \(\frac{{{x^2} – x}}{{x – y}}:\left( {{x^2} + xy} \right)\);

e) \(\left( {16 – {x^2}} \right):\left( {{x^2} – 4x} \right)\);                             

g) \(\frac{{4{y^2} – {x^2}}}{{{x^2} + 2xy + {y^2}}}:\frac{{x – 2y}}{{2{x^2} + 2xy}}\).

Lời giải chi tiết

a) \(\frac{{{x^2} – 5x}}{{4{y^2}}}:\frac{{5x}}{{2y}} = \frac{{x\left( {x – 5} \right)}}{{2.2y.y}}.\frac{{2y}}{{5x}} = \frac{{x\left( {x – 5} \right)2y}}{{2.2y.y.5x}} = \frac{{x – 5}}{{10y}}\)     ;                                    

b) \(\frac{{{x^2} – 1}}{y}:\frac{{x + 1}}{{{y^2}}} = \frac{{\left( {x – 1} \right)\left( {x + 1} \right)}}{y}.\frac{{y.y}}{{x + 1}} = \frac{{\left( {x – 1} \right)\left( {x + 1} \right).y.y}}{{y\left( {x + 1} \right)}} = y\left( {x – 1} \right)\);

c) \(\left( {{x^2} – 2xy} \right):\frac{{5x – 10y}}{x} = x\left( {x – 2y} \right).\frac{x}{{5\left( {x – 2y} \right)}} = \frac{{x\left( {x – 2y} \right).x}}{{5\left( {x – 2y} \right)}} = \frac{{{x^2}}}{5}\);                 

d) \(\frac{{{x^2} – x}}{{x – y}}:\left( {{x^2} + xy} \right) = \frac{{x\left( {x – 1} \right)}}{{x – y}}.\frac{1}{{x\left( {x + y} \right)}} = \frac{{x\left( {x – 1} \right)}}{{\left( {x – y} \right)x\left( {x + y} \right)}} = \frac{{x – 1}}{{{x^2} – {y^2}}}\);

e) \(\left( {16 – {x^2}} \right):\left( {{x^2} – 4x} \right) = \left( {4 – x} \right)\left( {4 + x} \right).\frac{1}{{x\left( {x – 4} \right)}} = \frac{{\left( {4 – x} \right)\left( {4 + x} \right)}}{{x\left( {x – 4} \right)}} = \frac{{ – x – 4}}{x}\); 

g) \(\frac{{4{y^2} – {x^2}}}{{{x^2} + 2xy + {y^2}}}:\frac{{x – 2y}}{{2{x^2} + 2xy}} = \frac{{\left( {2y – x} \right)\left( {2y + x} \right)}}{{{{\left( {x + y} \right)}^2}}}.\frac{{2x\left( {x + y} \right)}}{{x – 2y}}\)

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