若方程(x-1)(x^2-2x m)=0的三根是一个三角形三边的长-搜答案-百度作业网

(x-6)(x^2+x+1)-x(x+1)(x-1)=x(2-5x)x^3+x^2+x-6x^2-6x-6-x^3+x=2x-5x^2-4x-6=2x6x=-6x=-1

解分式方程：2x+1x

原分式方程化为：2x+1x(x+1)＝56(x+1)，方程的两边同乘6x（x+1），得6（2x+1）=5x，解得x=-67．检验：把x=-67代入6x（x+1）≠0，即x=-67是原分式方程的解．则原

x(2x-4)+3x(x-1)=5x(x-3)+82x²-4x+3x²-3x=5x²-15x+88x=8x=1

设x2+2x=y，则原方程为y-6y=1．解之得，y1=3，y2=-2．则x2+2x=3或x2+2x=-2，当x2+2x＝3时，有x2-3x+2=0，解之得，x1=1，x2=2．当x2+2x＝−2时，

2x(x-1)-x(3x+2)=-x(x+2)-122x^2-2x-3x^2-2x=-x^2-2x-122x^2-3x^2+x^2-2x-2x+2x=-12-2x+12=0-2(x-6)=0x-6=0

(x²-x+4)x-(x-1)(x²+2)=x+7x³-x²+4x-x³+x²-2x+2=x+72x+2=x+72x-x=7-2x=5

(x/x(x+2))+(x/(x+2)(x+4))+.+x/(x+8)(x+10)=(x/2)*(2/x(x+2))+(2/(x+2)(x+4))+.+2/(x+8)(x+10)=(x/2)*[1/x

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/

我给个思路吧（X+2-1/X+2)-(X+3-1/X+3)=（X+6-1/X+6)-(X+7-1/X+7)1-1/(X+2)-1+1/(X+3)=1-1/(X+6)-1+1/(X+7)剩下的就自己解决

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)

(X(13-X)(X+(13-X)))/(X+1)=42得分母是X^2(13-X)+(13-X)X分子是（X+1）再得分母X(13-X)(X+1)分子分母抵消的（X+1）X=42得X=6

通分化简得到(x+1)(x+2)=(x+5)(x+6)3x+2=11x+308x=-28x=-7/2

x=1再问：过程再答：再答：专门为你写的

即1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+9)-1/(x+10)=5/121/x-1/(x+10)=5/12两边乘12x(x+10)1

解方程：x+1x+2

方程变形得：x+2−1x+2-x+3−1x+3=x+6−1x+6-x+7−1x+7，即1-1x+2-1+1x+3=1-1x+6-1+1x+7，整理得：1x+3-1x+2=1x+7-1x+6，即x+2−

方程-1/3+x=2x!

x=-1/3再问：解释再答：

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

原方程变形为：x+1x+2-x+2x+3=x+7x+8-x+8x+9，去分母，得（x+2）（x+3）=（x+8）（x+9），整理，得12x+66=0，解得x=-5.5，经检验，x=-5.5是原方程的解

X-1/X-2-(X-2)/(X-3)=(X-5)/(X-6)-(X-6)/(X-7)(X-1)(X-3)-(X-2)²/（X-2）（x-3）=（X-5)(X-7)-(X-6)²/