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
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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
-x^2+ax-1
约分:(1) x^2+2x+1/x^2+x ; (2) 3x-6/x^2-4x+4; (3)ax^2/a+ax打错了应为::(1) (x^2+2x+1)/(x^2+x) ; (2) (3x-6)/(x^2-4x+4); (3)(ax^2)/(a+ax)
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函数f(x)=(x2+2x-3)/(x-1) (x>1) ax+1 (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
已知(ax^2+x-b)(ax^2-x+b)-ax(ax^3+x-1)的化简结果中不含x^2项和x项,求b^a
已知(2x+1)(3x-2)(ax*x+by+c)=6x*x*x*x+kx*x*x+lx*x-3x-2,分别求出k,l,a,b,c
(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^2-x-ax-1
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
已知{x|ax^2-ax+1
已知{x|ax^2-ax+1
集合{x|ax^2-ax+1
要使(ax^2-3x)(x^2-2x+1)的展开式中不含x^3项,则a=
要使(ax^2-3x)(x^2-2x-1)的展开式中不含x^3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=?
要使(ax²-3x)(x²-2x-1)的展开式中不含x³项,则a=