计算:{[-1/6x^(2n+1)y^(2n+1)+1/3x^(2n+1)y(^(2n-1)+1/2x^(2n-1)y^(2n-1)]}/[-1/12x^(2n-1)y^(2n-1)]
来源:百度知道 编辑:UC知道 时间:2024/06/01 02:36:52
计算:{[-1/6x^(2n+1)y^(2n+1)+1/3x^(2n+1)y(^(2n-1)+1/2x^(2n-1)y^(2n-1)]}/[-1/12x^(2n-1)y^(2n-1)]
过程!(^*__*^)
过程!(^*__*^)
(提公因式)
-1/6x^(2n+1)y^(2n+1)+1/3x^(2n+1)y(^(2n-1)+1/2x^(2n-1)y^(2n-1)
=-1/6x^(2n-1)y^(2n-1)[x^2y^2+2x^2+3x^(2n-1)]
所以
原式=-1/6x^(2n-1)y^(2n-1)[x^2y^2+2x^2+3x^(2n-1)]/[-1/12x^(2n-1)y^(2n-1)]
=2(x^2y^2-2x^2-3)
2(x^2y^2-2x^2-3)
1:分子上提取公因式:x^(2n-1)y^(2n-1)[-1/6x平方y平方+1/3x平方+1/2]
2:再比上分母得:2x平方y平方-4x平方-6
/是什么意思。
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