9x-5x=16解方程

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)(

原方程可变为：2x²-18x+16=0即,x²-9x+8=0把方程左边因式分解的,（x-1）（x-8）=0从而的,x-1=0或x-8=0解得,x=1或x=8

两边除以9^x(16/9)^x+(12/9)^x=1令a=(4/3)^x,即(12/9)^x=a(16/9)^x=[(4/3)^x]^2=a^2a^2+a-1=0a=(4/3)^x>0所以取正解a=(

9x^2+x=0把x提出来x(9x+1)=0x=0或9x+1=0所以x=0或-1/9

[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)=(

/>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/(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)

这道题不用那么复杂.先移项.(x-4)/(x-5)-(x-5)/(x-6)=(x-7)/(x-8)-(x-8)/( x-9)再通分.（x-4)(x-6)/(x-5)(x-6)-(x-5)(x

两边乘x(x+3)(x-3)2x³+3x²--27x-x²-3x=5x³-45x-3x³+2x²+27x-18-27x-3x=-45x+27

左右两边同乘以(X-5)(X-6)(X-8)(X-9)得到(X-4)(X-6)(X-8)(X-9)-(X-5)(X-5)(X-8)(X-9)=(X-5)(X-6)(X-7)(X-9)-(X-5)(X-

(x/x-5)-(4/x-5)=9（x-4)/(x-5)=9x-4=9x-458x=41x=41/8

(x-4)/(x-5)-(x-5)/(x-6)=(x-7)/(x-8)-(x-8)/(x-9)1+1/(x-5)-1-1/(x-6)=1+1/(x-8)-1-1/(x-9)1/(x-5)-1/(x-6

x-4/x-5-x-5/x-6=x-7/x-8-x-8/x-91+1/x-5-1-1/x-6=1+1/x-8-1-1/x-91/x-5-1/x-6=1/x-8-1/x-9-1/(x-5)(x-6)=-

5x+6x-9=2x11x-2x=99x=9x=1

(x-4)/(x-5)+(x-8)/(x-9)=(x-7)/(x-8)+(x-5)/(x-6)1+1/(x-5)+1+1/(x-9)=1+1/(x-8)+1+1/(x-6)1/(x-5)+1/(x-9

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