解方程x^4-3x^3 3x^2-2=0

4x^2+2x√(3x^2+x)+x-9=03x^2+x+2x√(3x^2+x)+x^2-9=0[√(3x^2+x)]^2+2x√(3x^2+x)+x^2=9[√(3x^2+x)+x]^2=9√(3x

乘出来3x³+5x²+5=-x³+4x³+5x²+x3x³+5x²+5=3x³+5x²+x两边减去3x

(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

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

((X-2)(X-4)-(X-3)^2)/(X-3)(X-4)=((X-5)(X-7)-(X-6)^2)/(X-6)(X-7),-1/(X^2-7X+12)=-1/(X^2-13X+42),6X=30

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

x^2-2x-3=0(x-3)(x+1)=0x=3,x=-1

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-

原式5/X（X-2）-4/X（X+2）=3/X²-45（X+2）/X(X-2)(X+2)-4(X-2)/X(X-2)(X+2)=3/X²-4(5X+10)/X-(4X-8)/X=3

(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)

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

通分得[（x-8）(x-4)-(x-9)(x-3)]/(x-3)(x-4)=[(x+7)(x+3)-(x+2)(x+8)]/(x+8)(x+3)解得5/(x-3)(x-4)=5/(x+8)(x+3),

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(2x-3)+3x+3x(x-2)=5x(x-4)+112x²-3x+3x+3x²-6x=5x²-20x+115x²-6x=5x²-20x+115x

移项得2x+3x-4x=18合并同类项得X=18

(x+8)/(x-3)-(x-9)/(x-4)=(x+7)/(x+8)-(x+2)/(x+3)[（x-8）(x-4)-(x-9)(x-3)]/(x-3)(x-4)=[(x+7)(x+3)-(x+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)+(

（2x-3)(-2x-3)+9x=x（3-4x)－（2x－3）（2x＋3）＋9x＝3x－4x²－（4x²－9）＋9x＝3x－4x²－4x²＋9＋9x＝3x－4x