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

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