已知(2x+xlgx)8的展开式中,二项式系数最大的项的值等于1 120,求x.

已知(2xxlgx)8的展开式中,二项式系数最大的项的值等于1 120,求x.

答案

C(2x)4(xlgx)41 120x4(1lgx)1,所以x1,或lgx=-1x.

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