解分式方程(5-2x)/(x^2-x-12)=5x/(x^2+x-6)+(10-7x)/(x^2-6x+8)得
2个回答
展开全部
(5-2x)/(x^2-x-12)=5x/(x^2+x-6)+(10-7x)/(x^2-6x+8)
(5-2x)/(x-4)(x+3)=5x/(x+3)(x-2)+(10-7x)/(x-2)(x-4)
分母同时乘以(x-4)(x+3)(x-2)得
(5-2x)(x-2)=5x(x-4)+(10-7x)(x+3)
即-2x^2+9x-10=5x^2-20x-7x^2-11x+30
9x-10=-31x+30
40x=40
x=1
(5-2x)/(x-4)(x+3)=5x/(x+3)(x-2)+(10-7x)/(x-2)(x-4)
分母同时乘以(x-4)(x+3)(x-2)得
(5-2x)(x-2)=5x(x-4)+(10-7x)(x+3)
即-2x^2+9x-10=5x^2-20x-7x^2-11x+30
9x-10=-31x+30
40x=40
x=1
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
展开全部
(5-2x)/(x^2-x-12)=5x/(x^2+x-6)+(10-7x)/(x^2-6x+8)
(5-2x)/{(x+3)(x-4)} = 5x/{(x+3)(x-2)} + (10-7x) / {(x-2)(x-4)}
(5-2x)/{(x+3)(x-4)} - 5x/{(x+3)(x-2)} - (10-7x) / {(x-2)(x-4)} = 0
{ (5-2x)(x-2) - 5x(x-4) - (10-7x)(x+3) } / {(x+3)(x-2)(x-4)} = 0
{ (-2x^2+9x-10) - (5x^2-20x) - (-7x^2-11x+30) } / {(x+3)(x-2)(x-4)} = 0
{ -2x^2+9x-10 - 5x^2 + 20x + 7x^2 +11x - 30) } / {(x+3)(x-2)(x-4)} = 0
40(x-1) / {(x+3)(x-2)(x-4)} = 0
x=1
(5-2x)/{(x+3)(x-4)} = 5x/{(x+3)(x-2)} + (10-7x) / {(x-2)(x-4)}
(5-2x)/{(x+3)(x-4)} - 5x/{(x+3)(x-2)} - (10-7x) / {(x-2)(x-4)} = 0
{ (5-2x)(x-2) - 5x(x-4) - (10-7x)(x+3) } / {(x+3)(x-2)(x-4)} = 0
{ (-2x^2+9x-10) - (5x^2-20x) - (-7x^2-11x+30) } / {(x+3)(x-2)(x-4)} = 0
{ -2x^2+9x-10 - 5x^2 + 20x + 7x^2 +11x - 30) } / {(x+3)(x-2)(x-4)} = 0
40(x-1) / {(x+3)(x-2)(x-4)} = 0
x=1
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
