Giải bài 3 trang 22 sách bài tập toán 8 - Chân trời sáng tạo

Đề bài

Thực hiện các phép cộng, trừ phân thức sau:

a) \(\frac{x}{{x + 2}} - \frac{x}{{x - 2}}\);           

b) \(\frac{{3x}}{{2y}} + \frac{{5x}}{{3y}}\);                         

c) \(\frac{{y - 1}}{{5y}} - \frac{{3x - 1}}{{15x}}\);

d) \(\frac{{1 - x}}{{{x^3}}} + \frac{1}{{{x^2}}}\);                 

e) \(\frac{{x - 2y}}{{x{y^2}}} - \frac{{y - 2x}}{{{x^2}y}}\);  

g) \(\frac{{1 - {y^2}}}{{3xy}} + \frac{{2{y^3} - 1}}{{6x{y^2}}}\).

Lời giải chi tiết

a) \(\frac{x}{{x + 2}} - \frac{x}{{x - 2}} = \frac{{x\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} - \frac{{x\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{{{x^2} - 2x - {x^2} - 2x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{{ - 4x}}{{{x^2} - 4}}\);   

b) \(\frac{{3x}}{{2y}} + \frac{{5x}}{{3y}} = \frac{{3x.3}}{{6y}} + \frac{{5x.2}}{{6y}} = \frac{{9x + 10x}}{{6y}} = \frac{{19x}}{{6y}}\); 

c) \(\frac{{y - 1}}{{5y}} - \frac{{3x - 1}}{{15x}} = \frac{{3x\left( {y - 1} \right)}}{{15xy}} - \frac{{y\left( {3x - 1} \right)}}{{15xy}} = \frac{{3xy - 3x - 3xy + y}}{{15xy}} = \frac{{y - 3x}}{{15xy}}\);

d) \(\frac{{1 - x}}{{{x^3}}} + \frac{1}{{{x^2}}} = \frac{{1 - x}}{{{x^3}}} + \frac{x}{{{x^3}}} = \frac{{1 - x + x}}{{{x^3}}} = \frac{1}{{{x^3}}}\);             

e) \(\frac{{x - 2y}}{{x{y^2}}} - \frac{{y - 2x}}{{{x^2}y}} = \frac{{x\left( {x - 2y} \right)}}{{{x^2}{y^2}}} - \frac{{y\left( {y - 2x} \right)}}{{{x^2}{y^2}}} = \frac{{{x^2} - 2xy - {y^2} + 2xy}}{{{x^2}{y^2}}} = \frac{{{x^2} - {y^2}}}{{{x^2}{y^2}}}\); 

g) \(\frac{{1 - {y^2}}}{{3xy}} + \frac{{2{y^3} - 1}}{{6x{y^2}}} = \frac{{2y\left( {1 - {y^2}} \right)}}{{6x{y^2}}} + \frac{{2{y^3} - 1}}{{6x{y^2}}} = \frac{{2y - 2{y^3} + 2{y^3} - 1}}{{6x{y^2}}} = \frac{{2y - 1}}{{6x{y^2}}}\).