4x²-7=12x的解

(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)(

化简：(1/x)*(1/(x+1))+(1/(x+1))*(1/(x+2))+(1/(x+2))*(1/(x+3))+(1/(x+3))*(1/(x+4))=4/21(1/x)-(1/(x+1))+(

((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

设a=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)那么y=a*(x-10);那么y^=a^*(x-10)+a*(x-10)^=a^*(x-10)+a那么y

[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^3+x^4+x^5+x^6+x^7+x^8=（x+x^2+x^3+x^4）+（x^5+x^6+x^7+x^8）=x（1+x+x^2+x^3）+x^5（1+x+x^2+x^3）=（x+x

【解法一】【解法二】5x(x+7)(x+4)=5x(x^2+11X)5x(x²+11x+28)-5x(x²+11x)=05x^3+55x^2+140x=5x^3+55x^25x(x

1+x+x^2+x^3=0x+x^2+x^3+x^4+x^5+x^6+x^7+x^8=(x+x^2+x^3+x^4)+(x^5+x^6+x^7+x^8)=x(1+x+x^2+x^3)+x^5(1+x+

通分得[（x-8）(x-4)-(x-9)(x-3)]/(x-3)(x-4)=[(x+7)(x+3)-(x+2)(x+8)]/(x+8)(x+3)解得5/(x-3)(x-4)=5/(x+8)(x+3),

题目应是1/(x^2+x﹚+1/(x^2+3x+2﹚+1/(x^2+5x+6﹚+1/(x^2+7x+12﹚=4/211/x(x+1﹚+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)

(x+8)/(x-3)-(x-9)/(x-4)=(x+7)/(x+8)-(x+2)/(x+3)[（x-8）(x-4)-(x-9)(x-3)]/(x-3)(x-4)=[(x+7)(x+3)-(x+2)(

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

首先由题意得x+1≠0,x+7≠0,x+5≠0,x+3≠0,即x≠-1,x≠-7,x≠-5,x≠-3,则先简化方程(x+1+1)/(x+1)+(x+7+1)/(x+7)=(x+5+1)/(x+5)+(

4x-60x+3x*x=5x-140x+7x*x4x*x-79x=0x(4x-79)=0x=0或x=79/4

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-

5x+2x=33.6,7x=33.6;x=4.8;7x-4x=0.75,3x=3/4;x=1/4;2x+7x+2.1=12,9x=8.1;x=0.9;56x-32x-119=36124x=480;x=

(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)=-

D