解方程x-2 x-x² -4＝1

6^x＋4^x＝9^x3^x*2^x+(2^x)^2=(3^x)^2,两边同除以(2^x)^2,（3/2）^x+1=[(3/2)^x]^2,设（3/2）^x=t,t+1=t^2,t^2-t-1=0,t

解x(6-x)-2x(x+1)-(x³-3x^4)/x²=46x-x²-2x²-2x-(x-3x²)=44x-3x²-x+3x²=

2x²-3x+2x²-2x=4x²-8x+9-5x=-8x+93x=9x=3

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~~

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

(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

1.x/x+1-1=3/（x+1)(x-2)两边乘以(x+1)(x-2)得x(x-2)-(x+1)(x-2)=3x²-2x-x²+x+2=3-x=1x=-1检验：x=-1是增根2.

题目应该是（x-1)(x-2)(x-3)(x-4)=24,不然解不了[(x-1)(x-4)][(x-2)(x-3)]=24[x^2-5x+4][x^2-5x+6]=24设x^2-5x+5=y所以(y-

(x-3)(x-2)-(x-4)(x-1))/((x-1)(x-2))=((x-5)(x-4)-(x-6)(x-3))/((x-3)(x-4))2/((x-1)(x-2))=2/((x-3)(x-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

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

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

（2x+5)(4x*x-10x+25)=8x(x*x+1)8x^3-20x^2+50x+20x^2-50x+125=8x^3+8x,8x=125,x=125/8或利用立方和公式8x^3+125=8x^

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)

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)+(

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

(3x+1)(3x-1)=(x+2)(9x-4)9x^2-1=9x^2+14x-8-14x=-7x=1/22(x-3)(x-2)-(x+3)²=(x-1)(x+1)2(x^2-5x+6)-(

(x-2)/(x+2)=(x+2)/(x-2)+16/(x²-4)方程两边同时乘(x²-4)(x-2)²=(x+2)²+16x²-4x+4=x

 再答：采纳再答：同学再答：有过程