高中数学：（1-x3「次方」）（1+x）10「次方」的展开式中，x5「次方」的...
\uff081-x3\uff08\u662f3\u6b21\u65b9\uff09\uff09\uff081+X)10\uff08\u662f10\u6b21\u65b9\uff09 \u4e0a\u5f0f\u5c55\u5f00\u5f0f\u4e2dX5\u6b21\u65b9\u7684\u7cfb\u6570\u662f\uff081-x^3)(1+x)^10 \u4e0a\u5f0f\u5c55\u5f00\u5f0f\u4e2dX5\u6b21\u65b9\u6765\u81ea1*(1+x)^10 \u4e2d\u5c55\u5f00\u7684\u542bx^5\u4e0e\uff08-x^3)*(1+x)^10 \u4e2d\u5c55\u5f00\u7684\u542bx^2\u5373C510+\uff08-1\uff09*C210
\u6839\u636e\u9898\u610f\uff0c\uff081+x3\uff09\uff081-x\uff0910\u7684\u5c55\u5f00\u5f0f\u4e2d\u6bcf\u4e00\u9879\u4e3a\uff081+x3\uff09\u4e2d\u7684\u4e00\u9879\u4e0e\uff081-x\uff0910\u7684\u5c55\u5f00\u5f0f\u4e2d\u4e00\u9879\u7684\u4e58\u79ef\uff0c\u800c\uff081-x\uff0910\u7684\u5c55\u5f00\u5f0f\u7684\u901a\u9879\u4e3aTr+1=C10r?\uff08-x\uff09r=\uff08-1\uff09rC10r?xr\uff0c\u8981\u5728\uff081+x3\uff09\uff081-x\uff0910\u7684\u5c55\u5f00\u5f0f\u51fa\u73b0x5\u9879\uff0c\u6709\u4e24\u79cd\u60c5\u51b5\uff0c\u2460\u3001\u82e5\uff081+x3\uff09\u4e2d\u51fa1\uff0c\u5219\uff081-x\uff0910\u4e2d\u5fc5\u987b\u51fax5\u9879\uff0c\u5219\u6b64\u65f6x5\u9879\u7684\u7cfb\u6570\u4e3a-C105\uff0c\u2461\u3001\u82e5\uff081+x3\uff09\u4e2d\u51fax3\u9879\uff0c\u5219\uff081-x\uff0910\u4e2d\u5fc5\u987b\u51fax2\u9879\uff0c\u5219\u6b64\u65f6x5\u9879\u7684\u7cfb\u6570\u4e3aC102\uff0c\u5219\u5728\uff081+x3\uff09\uff081-x\uff0910\u7684\u5c55\u5f00\u5f0f\u4e2d\uff0cx5\u7684\u7cfb\u6570\u662f-C105+C102=-252+45=-207\uff1b\u6545\u7b54\u6848\u4e3a-207\uff0e
既然是x5「次方」，而前面的式子为1-x3「次方」，你得把（1+x）10「次方」先展开此二项式部分，即需要(C10,2)(1^8)(X^2)与(C10,5).(1^5)(X^5)，(C10,2)(1^8)与x^3的系数相乘为-45，而(C10,5).(1^5)与1相乘则为252，故x^5的系数为207.希望你明白！
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