七年级的分式计算[(x+2)/2x][(x^3-3x^2)\(x^2+5x+6)]\[(x^3-9x)\(x^2+2x-
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/09/27 10:31:01
七年级的分式计算
[(x+2)/2x][(x^3-3x^2)\(x^2+5x+6)]\[(x^3-9x)\(x^2+2x-3)]
[(x^2-4y^2)/(x+y)]/(x^2-2xy)/[(x^2+xy-2y^2)\(x^2-y^2)]
[(x+2)/2x][(x^3-3x^2)\(x^2+5x+6)]\[(x^3-9x)\(x^2+2x-3)]
[(x^2-4y^2)/(x+y)]/(x^2-2xy)/[(x^2+xy-2y^2)\(x^2-y^2)]
1.
原式={[(x+2)/2x][x^2(x-3)/(x+2)(x+3)]}/{x(x+3)(x-3)/(x+3)(x-1)}
=[(x+2)(x-3)(x+3)(x-1)x^2]/[2x^2(x+2)(x+3)(x+3)(x-3)]
=(x-1)/2(x+3)
2.
原式=[(x+2y)(x-2y)(x+y)(x-y)]/[(x+y)(x-2y)(x+2y)(x-y)x^2]
=1/x^2
原式={[(x+2)/2x][x^2(x-3)/(x+2)(x+3)]}/{x(x+3)(x-3)/(x+3)(x-1)}
=[(x+2)(x-3)(x+3)(x-1)x^2]/[2x^2(x+2)(x+3)(x+3)(x-3)]
=(x-1)/2(x+3)
2.
原式=[(x+2y)(x-2y)(x+y)(x-y)]/[(x+y)(x-2y)(x+2y)(x-y)x^2]
=1/x^2
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