【(x^2-4分之x^2-x-6)+(x+2分之x-3)】除以(x+1分之x-3),帮帮忙明天就要交,
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/06 11:21:35
【(x^2-4分之x^2-x-6)+(x+2分之x-3)】除以(x+1分之x-3),帮帮忙明天就要交,
[(x^2-x-6)/(x^2-4)+(x-3)/(x+2)]÷(x-3)/(x+1)
=[(x-3)(x+2)/(x-2)(x+2)+(x-3)/(x+2)]÷(x-3)/(x+1)
=[(x-3)/(x-2)+(x-3)/(x+2)]÷(x-3)/(x+1)
=[(x-3)/(x-2)+(x-3)/(x+2)]*(x+1)/(x-3)
=(x-3)/(x-2)*(x+1)/(x-3)+(x-3)/(x+2)*(x+1)/(x-3)
=(x+1)/(x-2)+(x+1)/(x+2)
=(x+1)(x+2)/[(x-2)(x+2)]+(x+1)(x-2)/[(x-2)(x+2)]
=(x+1)[(x+2)+(x-2)]/[(x-2)(x+2)]
=2x(x+1)/[(x-2)(x+2)]
=[(x-3)(x+2)/(x-2)(x+2)+(x-3)/(x+2)]÷(x-3)/(x+1)
=[(x-3)/(x-2)+(x-3)/(x+2)]÷(x-3)/(x+1)
=[(x-3)/(x-2)+(x-3)/(x+2)]*(x+1)/(x-3)
=(x-3)/(x-2)*(x+1)/(x-3)+(x-3)/(x+2)*(x+1)/(x-3)
=(x+1)/(x-2)+(x+1)/(x+2)
=(x+1)(x+2)/[(x-2)(x+2)]+(x+1)(x-2)/[(x-2)(x+2)]
=(x+1)[(x+2)+(x-2)]/[(x-2)(x+2)]
=2x(x+1)/[(x-2)(x+2)]
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