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