若方程(x-1)(x*2-2x+m)=0
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(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
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
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−
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/