Đề bài

Thực hiện các phép cộng, trừ phân thức sau: a) \(\frac{b}{{a - b}} + \frac{{{a^2} - 3ab}}{{{a^2} - {b^2}}}\); b) \(\frac{{a + 3}}{{{a^2} - 1}} - \frac{1}{{{a^2} + a}}\); c) \(\frac{{2a}}{{{a^2} - 4a + 4}} + \frac{4}{{2 - a}}\); d) \(\frac{{a + 1}}{{{a^3} - 1}} - \frac{1}{{{a^2} + a + 1}}\).

Lời giải chi tiết

a) \(\frac{b}{{a - b}} + \frac{{{a^2} - 3ab}}{{{a^2} - {b^2}}} = \frac{{b\left( {a + b} \right)}}{{\left( {a - b} \right)\left( {a + b} \right)}} + \frac{{{a^2} - 3ab}}{{\left( {a - b} \right)\left( {a + b} \right)}} = \frac{{ab + {b^2} + {a^2} - 3ab}}{{\left( {a - b} \right)\left( {a + b} \right)}}\)\( = \frac{{{a^2} - 2ab + {b^2}}}{{\left( {a - b} \right)\left( {a + b} \right)}} = \frac{{{{\left( {a - b} \right)}^2}}}{{\left( {a - b} \right)\left( {a + b} \right)}} = \frac{{a - b}}{{a + b}}\)b) \(\frac{{a + 3}}{{{a^2} - 1}} - \frac{1}{{{a^2} + a}} = \frac{{a\left( {a + 3} \right)}}{{a\left( {a - 1} \right)\left( {a + 1} \right)}} - \frac{{a - 1}}{{a\left( {a - 1} \right)\left( {a + 1} \right)}} = \frac{{{a^2} + 3a - a + 1}}{{a\left( {a - 1} \right)\left( {a + 1} \right)}}\)\( = \frac{{{a^2} + 2a + 1}}{{a\left( {a - 1} \right)\left( {a + 1} \right)}} = \frac{{{{\left( {a + 1} \right)}^2}}}{{a\left( {a - 1} \right)\left( {a + 1} \right)}} = \frac{{a + 1}}{{a\left( {a - 1} \right)}}\)c) \(\frac{{2a}}{{{a^2} - 4a + 4}} + \frac{4}{{2 - a}} = \frac{{2a}}{{{{\left( {a - 2} \right)}^2}}} - \frac{{4\left( {a - 2} \right)}}{{{{\left( {a - 2} \right)}^2}}} = \frac{{2a - 4a + 8}}{{{{\left( {a - 2} \right)}^2}}} = \frac{{ - 2a + 8}}{{{{\left( {a - 2} \right)}^2}}}\); d) \(\frac{{a + 1}}{{{a^3} - 1}} - \frac{1}{{{a^2} + a + 1}} = \frac{{a + 1}}{{\left( {a - 1} \right)\left( {{a^2} + a + 1} \right)}} - \frac{{a - 1}}{{\left( {a - 1} \right)\left( {{a^2} + a + 1} \right)}} = \frac{{a + 1 - a + 1}}{{\left( {a - 1} \right)\left( {{a^2} + a + 1} \right)}}\)\( = \frac{2}{{{a^3} - 1}}\)