要使（x^2+ax+5)(-6x^3+x)的展开式中不含有x^4项,则a=?

要使（x^2+ax+5)(-6x^3+x)的展开式中不含有x^4项,则a=? *-----------------------------------------------*| 6 4 X | 8 X X | X X 5 || X X X | X X X | X 7 8 || X X X | X X X | X X X ||---------------+---------------+--------------- || X X X | X X X | 5 1 X || X X X | X 6 X | X X X || 8 X X | 3 5 X | 2 X X || 填九宫格帮帮忙.x x 6 x x 7 x x 98 x x x 3 x 1 x x 9 x x 6 x 5 x 3 x x x 3 x x x x 1 8x x x 9 x 1 x x x2 1 x x x x 6 x x x 6 x 7 x 3 x x 1 x x 9 x 2 x x x 47 x x 8 x x 5 x x 要使(x^2+ax+5)(-6x^3)的展开式中不含x^4项,则a应等于 (x+1)(x+2)(x+3)(x+4)(x+5)(x+6)(x+7)(x+8)(x+9)(x+10)请化简我说的是化简化成x^10+ax^9+bx^8+cx^7+......+n的形式 因式分解：x^3+6x^3+11x+6 6x^4+5x^3+3x^2-3x-2 x^4-11x^2+1 2x^3-ax^2-5a^2x-2a^3x^3+6x^2+11x+66x^4+5x^3+3x^2-3x-2x^4-11x^2+12x^3-ax^2-5a^2x-2a^3 要使(x^2+ax+1)(-6x^3+x^4)的展开式中不含X^4项,则a应等于 ax已知x·x+3x-2=0,求2·x·x·x+6·x·x-4x的值. 2x^5-3x^3x√3+√23a^6x^3-9ax^5 2x²-5x-ax+3a-3 因式分解 分解因式：2x^3+x-3还有x^2+2x-15-ax-5a (x*x+ax+b)(2x*x-3x-1)的积中,x*x*x项的系数为-5,x*x项的系数为-6 求a,b的值 若整事(2x^2+3ax-x+4)-(3x^2-5ax+2x)+(x^2-ax+6)的值与x无关,试确定a的值 有这样一道题：若整式（2x^2+3ax-x+4)-(3x^2-5ax+2x)+(x^2-ax+6)的值与字母x无关,是确定a的值 因式分解：①2ax^2-11ax+5x ②3x^3+3x^2-18x 若整式（2x²+3ax-x+4）-（3x²-5ax+2x）+（x²-ax+6）的值与x无关,试确定a的值 若整式(2x²+3ax-x+4)-(3x²-5ax+2x)+(x²-ax+6)的值与字母x无关,试确定a的值 不等式x^2+ax-6