若方程4x-6=1-2x和8-k=2x 2的解相同

(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

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

(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/

解方程：4x+2m=3x+1,移项：4x-3x=1-2m,合并同类项：x=1-2m解方程：3x+2m=6x+1,移项：3x-6x=1-2m,合并同类项：-3x=1-2m,系数化1：x=(1-2m)÷(

x+1分之x+2减去x+3分之x+4=x+5分之x+6减去X+7分之x+8方程两边分别通分,得[(x+2)(x+3)-(x+1)(x+4)]/[(x+1)(x+3)]=[(x+6)(x+7)-(x+8

首先整理下7X-15/X-4=14X-31/2X-8,31/2X-15/X=7X-4,两边都乘以2X得14X*X-8X-1=0.然后根据公式X=-b正负根号(§)下b方-4ac得解为X1=(4+§30

1、求出两方程的解x=1-2m和x=(2m-1)/3,两解相同,即1-2m=(2m-1)/3求出m=1/22、代入第一项为0,第二项为1,即得0-1=-1

(x-9)/(x-7)-(x+1)/(x-1)=(x-8)/(x-6)-x/(x-2)[(x-9)(x-1)-(x+1)(x-7)]/[(x-7)(x-1)]=[(x-8)(x-2)-x(x-6)]/

2x+4x+6x...+100x=1-(x+3x+5x+...+99x)x(2+4+6+...+100)=1-x(1+3+5+...99)x(2+4+6+...+100)+x(1+3+5+...+99

（1）由6x+2a-1=5x得x=-2a+1①由4x+2a=7x+1得x=2a−13②∴-2a+1=2a−13解得：a=12（2）将a=12代入：原式=(12+3)2009•(2×12−97)2010

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/

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)+(

2x+4x+6x+8x+...+2010x+2012x=1013042x+2x+3x+4x+...+1005x+1006x=5065211006*1007x/2=506521x=1013042/100

等号前后同时分别计算减法得：(x+1)(x+3)分之2=(x+5)(x+7)分之2所以：(x+1)(x+3)=(x+5)(x+7)x^2+4x+3=x^2+12x+35x=-4

2x+4x+6x+8x~+2010x+2012x=1013042（2+2012)×1006÷2×x=1013042x=1再问：可以分解仔细点吗？再答：（2+2012)×1006÷2×x=1013042

1,应该是x是几次幂就有几个根吧,所以应该是7+2=9个2.-7.036622145,-6.489288572,-6.200988153,-4.734373320,-4.289168546,-2.77

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/

分别代入,看等式是否成立判断0：3（2x+1）+6x=39x+5（x-1）=-5等式不成立,不是方程的解判断4：3（2x+1）+6x=3（8+1）+24=27+24=519x+5（x-1）=36+5（

等号右边的第一个x+3应为x+5吧?拆项：1+1/(x+1)+1+1/(x+7)=1+1/(x+5)+1+1/(x+3)1/(x+1)+1/(x+7)=1/(x+5)+1/(x+3)通分：(2x+8)