方程x^3-6x² 9x-10=0的实根有

判断方程的元数只要看方程中有几个未知数就可以了,显然此方程是一元,又因为未知数x还有做除数的而且有多个等式,所以此方程是一元二次方程组.楼主应该可以自行解决吧

(x+3)/(x+2)+(x+9)/(x+8)=(x+5)/(x+4)+(x+7)/(x+6)1+(x+5)/(x+2)(x+8)=1+(x+5)/(x+4)(x+6)x=-5再问：第二步那个1是怎么

将x-7/x-9分解成1+2/(x-9)其他分式同理则原方程等价于1/x-9+1/x-5=1/x-6+1/x-81/x-6-1/x-5=1/x-9-1/x-81/(x-5)(x-6)=1/(x-8)(

((X-2)(X-4)-(X-3)^2)/(X-3)(X-4)=((X-5)(X-7)-(X-6)^2)/(X-6)(X-7),-1/(X^2-7X+12)=-1/(X^2-13X+42),6X=30

(x-3-1)/(x-3)-(x-5-1)/(x-5)=(x-7-1)/(x-7)-(x-9-1)/(x-9)1-1/(x-3)-1+1/(x-5)=1-1/(x-7)-1+1/(x-9)1/(x-7

[1-1/(x-4)]+[1-1/(x-8)]=[1-1/(x-7)]+[1-1/(x-5)]1/(x-4)+1/(x-8)=1/(x-7)+1/(x-5)(x-4+x-8)/(x-4)(x-8)=(

x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)1+2/(x-2)+1-2/(x-7)=1+2/(x-1)+1-2/(x-6)1/(x-2)-1/(x-7)=1/

方程实根个数就是曲线y=x^3-6x^2+9x-10与x轴的交点个数.y'=3x^2-12x+9=>y'(3)=0,y'(1)=0y(3)=-10,y（1）=4∵y(x->-∞）->-∞,y(1)=4

/>因为：x-5/x-6+x-8/x-9=x-7/x-8+x-6/x-7,所以：1+1/(x-6)+1+1/(x-9)=1+1/(x-8)+1+1/(x-7)即：1/(x-6）+1/(x-9)=1/(

(1)(X+7)/(X+6)+(X+9)/(X+8)=(X+10)/(X+9)+(X+6)/(X+5)这类方程重在运算的技巧性,观察分母7+9=10+6;6+8=5+9所以先移项:(X+7)/(X+6

(x-9)/(x-7)-(x+1)/(x-1)=(x-8)/(x-6)-x/(x-2)[(x-9)(x-1)-(x+1)(x-7)]/[(x-7)(x-1)]=[(x-8)(x-2)-x(x-6)]/

x/(x-2)=2x/(x-3)+(1-x)/(x-5x+6)x/(x-2)=2x/(x-3)+(1-x)/(x-2)(x-3)x(x-3)/(x-2)(x-3)=2x(x-2)/(x-2)(x-3)

6x^2-6x^2-4x+9x=-105x=-10x=-2

自己算吧

3x=-6x3x+6x=09x=0x=0

移项3x-x=102x=10x=10÷2x=5再问：还有一题：翟志刚如果在太空中行走20秒钟，将会飞过156千米，照这样计算，他这次在太空中共行驶了1125秒钟，共飞行了多少千米？再答：156÷20×

1,应该是x是几次幂就有几个根吧,所以应该是7+2=9个2.-7.036622145,-6.489288572,-6.200988153,-4.734373320,-4.289168546,-2.77

你好,我认真解答了这道题,看最终没有实数解啊.我希望能帮到你

解方程：(x-2)(x^2-6x-9)=x(x-5)(x-3)将方程展开：X^3-6x^2-9x-2x^2+12x+18=x^3-8x^2+15x合并同类项得：12x=18则x=18/12=3/2