Đề bài

Tính các tổng sau: a) \(\frac{5}{{6{x^2}y}} + \frac{7}{{12x{y^2}}} + \frac{{11}}{{18xy}};\) b) \(\frac{{{x^3} + 2x}}{{{x^3} + 1}} + \frac{{2x}}{{{x^2} - x + 1}} + \frac{1}{{x + 1}}\)

Lời giải chi tiết

a) \(\frac{5}{{6{x^2}y}} + \frac{7}{{12x{y^2}}} + \frac{{11}}{{18xy}} \\= \frac{{5.6y}}{{36{x^2}{y^2}}} + \frac{{7.3x}}{{36{x^2}{y^2}}} + \frac{{11.2xy}}{{36{x^2}{y^2}}} \\= \frac{{30y + 21x + 22xy}}{{36{x^2}{y^2}}}\)b)\(\frac{{{x^3} + 2x}}{{{x^3} + 1}} + \frac{{2x}}{{{x^2} - x + 1}} + \frac{1}{{x + 1}} \\= \frac{{{x^3} + 2x}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} + \frac{{2x\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} + \frac{{{x^2} - x + 1}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\)\( \\= \frac{{{x^3} + 2x + 2{x^2} + 2x + {x^2} - x + 1}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} \\= \frac{{{x^3} + 3{x^2} + 3x + 1}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} \\= \frac{{{{\left( {x + 1} \right)}^3}}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} \\= \frac{{{x^2} + 2x + 1}}{{{x^2} - x + 1}}\)