Thực hiện phép tính:
a) \(\frac{1}{{x + 1}} + \frac{2}{{1 - x}} + \frac{{5x - 1}}{{{x^2} - 1}}\)
b) \(\frac{{2x + 6}}{{{x^3} - 8}}:\frac{{{{\left( {x + 3} \right)}^3}}}{{2x - 4}}\)
Sử dụng các quy tắc tính với phân thức để thực hiện phép tính.
a) \(\frac{1}{{x + 1}} + \frac{2}{{1 - x}} + \frac{{5x - 1}}{{{x^2} - 1}}\) (ĐK: \(x \ne \pm 1\))
\( = \frac{1}{{x + 1}} - \frac{2}{{x - 1}} + \frac{{5x - 1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\)
\(\begin{array}{l} = \frac{{x - 1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} - \frac{{2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} + \frac{{5x - 1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\\ = \frac{{x - 1 - 2\left( {x + 1} \right) + 5x - 1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\\ = \frac{{x - 1 - 2x - 2 + 5x - 1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\\ = \frac{{4x - 4}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\\ = \frac{{4\left( {x - 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\\ = \frac{4}{{x + 1}}\end{array}\)
b) \(\frac{{2x + 6}}{{{x^3} - 8}}:\frac{{{{\left( {x + 3} \right)}^3}}}{{2x - 4}}\) (ĐK: \(x \ne 2\))
\(\begin{array}{l} = \frac{{2\left( {x + 3} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}:\frac{{{{\left( {x + 3} \right)}^3}}}{{2\left( {x - 2} \right)}}\\ = \frac{{2\left( {x + 3} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}.\frac{{2\left( {x - 2} \right)}}{{{{\left( {x + 3} \right)}^3}}}\\ = \frac{4}{{\left( {{x^2} + 2x + 4} \right){{\left( {x + 3} \right)}^2}}}\end{array}\)