5x/(x^2+x-6)+(2x-5)/(x^2-x-12)=(7x-10)/(x^2-6x+8)
来源:百度知道 编辑:UC知道 时间:2024/05/10 18:01:04
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/前是分子,/后是分母
/前是分子,/后是分母
5x/(x^2+x-6)+(2x-5)/(x^2-x-12)=(7x-10)/(x^2-6x+8)
5x/[(x+3)(x-2)]+(2x-5)/[(x-4)(x+3)]=(7x-10)/[(x-2)(x-4)]
同时乘以(x+3)(x-2)(x-4)
得到 5x(x-4)+(2x-5)(x-2)=(7x-10)(x+3)
5x^2-20x+2x^2-9x+10=7x^2+11x-30
40x=40
x=1
5x/[(x+3)(x-2)]+(2x-5)/[(x-4)(x+3)]=(7x-10)/[(x-2)(x-4)]
同时乘以(x-2)(x+3)(x-4)
5x(x-4)+(2x-5)(x-2)=(7x-10)(x+3)
5x^2-20x+2x^2-9x+10=7x^2+11x-30
40x=40
x=1
(x^2+2x-3)/(x^2-9)*(x^2-5x+6)/(3x^2-x-2)
5x/(x^2+x-6)+(2x-5)/(x^2-x-12)=(7x-10)/(x^2-6x+8)
(X-1/X)=5,且X<0,求x^10+x^6+x^4+1除以x^10+x^8+x^2+1的值
(x^2-x)(x^2-5x+6)<0
分式方程问题:解方程:[x+1)/(x+2)]+[(x+6)/(x+7)]=[(x+2)/(x+3)]+[(x+5)/(x+6)]
(x+2/x+1)+ (x+6/x+5)=(x+3/x+2)+(x+5/x+4)怎么解?
x-5/x-7+x-2/x--4=x-3/x-5+x-4/x-6
x^2-3x+1=0,求2x^5-5x^4-x^3+x^2-6x/x^2+1
方程(x+2)/(x+1)-(x+4)/(x+3)=(x+6)/(x+5)-(x+8)/(x+7)
[x+2]/[x+1]-[x+4]/[x+3]-[x+3]/[x+2]+]x+5]/[x+4]