已知(x²+nx+3)(x²﹣3x+m)的展开式中不含x²和x³项,则m=______,n=______.求解释、
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![已知(x²+nx+3)(x²﹣3x+m)的展开式中不含x²和x³项,则m=______,n=______.求解释、](/uploads/image/z/1261306-10-6.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%88x%26%23178%3B%EF%BC%8Bnx%EF%BC%8B3%EF%BC%89%EF%BC%88x%26%23178%3B%EF%B9%A33x%EF%BC%8Bm%EF%BC%89%E7%9A%84%E5%B1%95%E5%BC%80%E5%BC%8F%E4%B8%AD%E4%B8%8D%E5%90%ABx%26%23178%3B%E5%92%8Cx%26%23179%3B%E9%A1%B9%2C%E5%88%99m%EF%BC%9D______%2Cn%EF%BC%9D______.%E6%B1%82%E8%A7%A3%E9%87%8A%E3%80%81)
已知(x²+nx+3)(x²﹣3x+m)的展开式中不含x²和x³项,则m=______,n=______.求解释、
已知(x²+nx+3)(x²﹣3x+m)的展开式中不含x²和x³项,则m=______,n=______.
求解释、
已知(x²+nx+3)(x²﹣3x+m)的展开式中不含x²和x³项,则m=______,n=______.求解释、
(x²+nx+3)(x²﹣3x+m)
=x⁴-3x³+mx²+nx³-3nx²+mnx+3x²-9x+3m
=x⁴+(n-3)x³+(m-3n+3)x²+(mn-9)x+3m
∵展开式中不含x²和x³项
∴n-3=0,m-3n+3=0
∴n=3,m=6
(x²+nx+3)(x²﹣3x+m)的展开式中x²项为:(m-3n+3)x²; x³项为:(-3+n)x³
(x²+nx+3)(x²﹣3x+m)的展开式中不含x²和x³项,则m-3n+3=0,(-3+n)=0
=>n=3,m=6