MathematicsEasy64×since 2002Q2921If the third term in the binomial expansion of (1+xlog2x)5{\left( {1 + {x^{{{\log }_2}x}}} \right)^5}(1+xlog2x)5 equals 2560, then a possible value of x is -A222\sqrt 222B424\sqrt 242C18{1 \over 8}81D14{1 \over 4}41Check answerSkip