Đề bài

Rút gọn các phân thức sau: a)     \(\frac{{24{a^5}{b^3}}}{{18{a^3}{b^4}}}\) b)    \(\frac{{2x - {x^2}}}{{{x^2}y - 4y}}\) c)     \(\frac{{12{x^2} + 28x + 8}}{{9{x^2} - 1}}\) d)    \(\frac{{{x^3} + {x^2} + x + 1}}{{{x^2} - 1}}\)

Lời giải chi tiết

a)     \(\frac{{24{a^5}{b^3}}}{{18{a^3}{b^4}}} = \frac{{6{a^3}{b^3}.4{a^2}}}{{6{a^3}{b^3}.3{b^2}}} = \frac{{4{a^2}}}{{3{b^2}}}\)b)    \(\frac{{2x - {x^2}}}{{{x^2}y - 4y}} = \frac{{x\left( {2 - x} \right)}}{{y\left( {{x^2} - 4} \right)}} = \frac{{ - x\left( {x - 2} \right)}}{{y\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{{ - x}}{{y\left( {x + 2} \right)}}\)c)     \(\frac{{12{x^2} + 28x + 8}}{{9{x^2} - 1}} = \frac{{\left( {3x + 1} \right)\left( {x + 2} \right)}}{{\left( {3x - 1} \right)\left( {3x + 1} \right)}} = \frac{{x + 2}}{{3x - 1}}\)d)    \(\frac{{{x^3} + {x^2} + x + 1}}{{{x^2} - 1}} = \frac{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{{{x^2} + 1}}{{x - 1}}\)