Khai triển nhị thức \({\left( {2{x^2} - \frac{1}{2}} \right)^5}\) ta được số hạng chứa \({x^6}\) là:
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Ta có
\(\begin{align} {\left( {2{x^2} - \frac{1}{2}} \right)^5}& = {\left( {2{x^2}} \right)^5} - 5.{\left( {2{x^2}} \right)^4}.\frac{1}{2} + 10.{\left( {2{x^2}} \right)^3}.{\left( {\frac{1}{2}} \right)^2} - 10.{\left( {2{x^2}} \right)^2}.{\left( {\frac{1}{2}} \right)^3} + 5.\left( {2{x^2}} \right).{\left( {\frac{1}{2}} \right)^4} - {\left( {\frac{1}{2}} \right)^5}\\ &= 32{x^{10}} - 40{x^8} + 20{x^6} - 5{x^4} + \frac{5}{8}{x^2} - \frac{1}{{32}}\end{align}\).
Suy ra số hạng chứa \({x^6}\) là \(20{x^6}\).
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