解方程.(5x/x²+x-6)+(2x-5/x²-x-12)=7x-10/x²-6x+8

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/02 14:26:23

解方程.(5x/x²+x-6)+(2x-5/x²-x-12)=7x-10/x²-6x+8
解方程.(5x/x²+x-6)+(2x-5/x²-x-12)=7x-10/x²-6x+8

解方程.(5x/x²+x-6)+(2x-5/x²-x-12)=7x-10/x²-6x+8
5x/(x²+x-6)+(2x-5)/(x²-x-12)=(7x-10)/(x²-6x+8)
5x/[(x+3)(x-2)]+(2x-5)/[(x-4)(x+3)]=(7x-10)/[(x-2)(x-4)]
通分化简,得
5x(x-4)+(2x-5)(x-2)=(7x-10)(x+3)
整理,得
40x=40
x=1
代回原方程检验,分母有意义,x=1是方程的解.
x=1