Tìm số hạng chứa \({x^5}\) trong khai triển \({\left( {{x^2} + \frac{1}{x}} \right)^4}\)

Đáp án đúng là C

Ta có \({\left( {{x^2} + \frac{1}{x}} \right)^4} = C_4^0{\left( {{x^2}} \right)^4} + C_4^1{\left( {{x^2}} \right)^3}.\left( {\frac{1}{x}} \right) + C_4^2{\left( {{x^2}} \right)^2}.{\left( {\frac{1}{x}} \right)^2} + C_4^3\left( {{x^2}} \right).{\left( {\frac{1}{x}} \right)^3} + C_4^4{\left( {\frac{1}{x}} \right)^4}\)

\( = C_4^0{x^8} + C_4^1{x^6}.\left( {\frac{1}{x}} \right) + C_4^2{x^4}.\frac{1}{{{x^2}}} + C_4^3\left( {{x^2}} \right).\left( {\frac{1}{{{x^3}}}} \right) + C_4^4\left( {\frac{1}{{{x^4}}}} \right) = {x^8} + 4{x^5} + 6{x^2} + \frac{4}{x} + \frac{1}{{{x^4}}}\).