Câu hỏi
Khai triển nhị thức \(P\left( x \right) = {\left( {{x^2} + \frac{1}{2}} \right)^5}\) ta được
A
\(C_5^0.{x^{10}} - C_5^1.{x^8}.\frac{1}{2} + C_5^2.{x^6}.{\left( {\frac{1}{2}} \right)^2} - C_5^3.{x^4}.{\left( {\frac{1}{2}} \right)^3} + C_5^4.{x^2}.{\left( {\frac{1}{2}} \right)^4} - C_5^5.{\left( {\frac{1}{2}} \right)^5}\).
B
\(C_5^0.{x^{10}} + C_5^1.{x^8}.2 + C_5^2.{x^6}{.2^2} + C_5^3.{x^4}{.2^3} + C_5^4.{x^2}{.2^4} + C_5^5{.2^5}\).
C
\(C_5^0.{x^{10}} + C_5^1.{x^8}.\frac{1}{2} + C_5^2.{x^6}.{\left( {\frac{1}{2}} \right)^2} + C_5^3.{x^4}.{\left( {\frac{1}{2}} \right)^3} + C_5^4.{x^2}.{\left( {\frac{1}{2}} \right)^4} + C_5^5.{\left( {\frac{1}{2}} \right)^5}\).
D
\(C_5^0.{x^{10}} - C_5^1.{x^8}.2 + C_5^2.{x^6}{.2^2} - C_5^3.{x^4}{.2^3} + C_5^4.{x^2}{.2^4} - C_5^5{.2^5}\).
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