已知X-1分之X=Y 4Y-1分之

（1）y=x/x(x-1)*(x-1)²/(x-1)(x+1)-2/(x+1)=1/(x+1)-2/(x+1)=-1/(x+1)=1/3∴x+1=-3,x=-4即x=-4时,y=1/3（2）

A/(x-3)+1/(x+4)=[A(x+4)+(x-3)]/(x-3)(x+4)=(Ax+4A+x-3)/(x-3)(x+4)=[(A+1)x+(4A-3)]/(x-3)(x+4)∵(2x+1)/(

1/X(X+1)+1/(x+1)(x+2)+.+1/(x+99)(x+100)=1/x-1/(x+1)+1/(x+1)-1/(x+2)+.+1/(x+99)-1/(x+100)=1/x-1/(x+10

2-（X-1）/3=(1-X)/2+3-X(7-X)/3=(7-3X)/214-2X=21-9X7X=7X=1把X=1代入方程得4-K/3=0K=12

原式=[(1-1分之x+5)-(1-1分之x+4))+(1-1分之x+3)-(1-1分之x+2)]*[(x+3)(x+5)]\(x^2+7x+13)=[(1分之x+2)-(1分之x+3)+(1分之x+

已知分式4x^2-1分之6x^2+x-2的值为0,则x^-1=∵分式4x^2-1分之6x^2+x-2的值为0∴6x²+x-2=04x²-1≠0(3x+2)(2x-1)=0∴3x+2

由于A/(X+6)+B/(X-3)=(AX-3A+BX+6B)/(X+6)(X-3)=(2X+1)/(x+6)(x-3)所以,A+B=2,6B-3A=1,解得,A=11/9B=7/9

[(x+2)+1]/(x+2)-[(x+1)+1]/(x+1)=[(x+4)+1]/(x+4)-[(x+3)+1]/(x+3)1+1/(x+2)-1-1/(x+1)=1+1/(x+4)-1-1/(x+

(x²-x+1)/x=5x-1+1/x=5x+1/x=6那么(x^4+x²+1)/x²=x²+1+1/x²=(x+1/x)²-1=6

x/(x^2-x+1)=1x^2-x+1=xx^2+1=2x两边平方x^4+2x^2+1=4x^2两边减去x^2x^4+x^2+1=3x^2所以x^2/(x^4+x^2+1)=1/3

(x-1/x)^2=(x+1/x)^2-4=(√7)^2-4=7-4=3x-1/x=±√3

x/(x^2+x+1)=-1/2即-2x=x^2+x+1x^2+3x+1=0(x-1)/(2x^2+x^3-1)/(x^2-x+1)=(x-1)/(2x^2+x^3-1)(x^2-x+1)=(x-1)

x+y+z=0x+y=-zx+z=-yy+z=-x(x+y+z)^3=0x^3+y^3+z^3+3x^2y+3x^2z+3y^2x+3y^2z+3z^2x+3z^2y+6xyz=0x^3+y^3+z^

x²-2分之x²=33x²-6=x²2x²=6x²=3(1-x分之1-1+x分之1)除以（x²-1分之x+x)=[(1-x)(1+

设4x^2=a,4y^2=b,4z^2=b,则有x=二分之一根号a,y=二分之一根号b,z=二分之一根号c,将abc代入式子中得：

不用也可以算(x^4+x^2+1)/x^2=x^4/x^2+x^2/x^2+1/x^2=x^2+1+1/x^2你说的不对这里没有x+1/x而是x^2+2+1/x^2这个是完全平方即x^2+2*x*1/

x-1/x=3两边平方x^2-2+(1/X)^2=9x^2+(1/X)^2=11

等于负5分之4乘以根号5

x²+x²分之1=14（x+1/x）²-2=14（x+1/x）²==16x+1/x=±4如果明白,并且解决了你的问题,

数学之美团为你解答3x/(x-4y)+(x+y)/(4y-x)-7x/(x-4y)=3x/(x-4y)-(x+y)/(x-4y)-7x/(x-4y)=(3x-x-y-7x)/(x-4y)=(-5x-y