解方程 2x²+7x+1＝0

两边乘以(x+1)(x-1)得x²-3x+(2x-1)(x+1)=0x²-3x+2x²+x-1=03x²-2x-1=0(3x+1)(x-1)=0∴x=-1/3x

2（x-2)(x+3)-(2x-5)(x-1)=3x+72（x²+x-6）-（2x²-7x+5）=3x+72x²+2x-12-2x²+7x-5=3x+79x-1

4/(x-2)+(x-1)/[(x-2)(x-3)]-2/(x-3)=0[4(x-3)+(x-1)-2(x-2)]/[(x-2)(x-3)]=03(x-3)/[(x-2)(x-3)]=0不知道解了~~

(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

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

2x²-x-3-x²-x+2=x²+7x²-2x-1=x²+72x=-8x=-4

方程两边同时乘以x²-1:(3x²+9x+7)(x-1)-(2x²+4x-3)(x+1)-(x³+x+1)=03x^3+9x^2+7x-3x^2-9x-7-(2

(x+1)(x+2)(x^2-2x-1)(x-3)(x-4)+24=0(x+1)(x-3)(x+2)(x-4)(x^2-2x-1)+24=0(x^2-2x-3)(x^2-2x-8)(x^2-2x-1)

2/x^2+x+3/x^2-x-4/x^2-1=0(2/x^2+3/x^2-4/x^2)+x-x-1=01/x^2-1=01/x^2=1x^2=1x=1或-1

7/(x^2+x)+3/(x^2-x)=6/(x^2-1)7/x(x+1)+3/x(x-1)=6/(x-1)(x+1)分子分母同乘以x(x-1)(x+1),得7（x-1）+3（x+1)=6x10x-4

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

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)+(

3x^2-7x＝0x²-7x/3=0(x-7/3)x=0x=0,x=7/3

x(2x-5)-2(x-1)(x+7)=02x²-5x-2(x²+6x-7)=02x²-5x-2x²-12x+14=0-17x=-14x=14/17.

这个展开就行了哇.2*（x^2+x-6）-(2*x^2-7X+5)-3x-7=0化简得：6X=24x=4应该是对的.嘿嘿.

7/x(x+1)-6/(x+1)(x-1)+3/x(x-1)=0两边乘x(x+1)(x-1)7(x-1)-6x+3(x+1)=07x-7-6x+3x+3=04x=4x=1经检验,x=1时公分母x(x+

（x+1/x)^2-{7(x^2+1)/2x}+5/2=02（x+1/x)^2-7（x+1/x)+5=0（x+1/x)=1或5/2x^2-x+1=0无解2x^2-5x+2=0x=2或1/2

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/

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

再答：你可以问问你们老师，，，这样的话分母为0，那么还有意义吗