若方程x²-3x-1=0-搜答案-大师作文网

答：题目应该是有错误吧?(x²+1)/(x-1)-(3x-3)/(x²+1)=0设a=(x²+1)/(x-1),原方程化为：a-3/a=0所以：a=3/aa^2=3解得：

(2x+1)(3x-2)-6(x+1)(x-1)=06x^2-4x+3x-2-6(x^2-1)=06x^2-x-2-6x^2+6=0-x+4=0x=4再问：为什么我做的最后对于3/7再答：你怎么做的呀

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

x-1分之3-x(x-1)分之x+2=03x-(x+2)=02x=2x=1检验：x=1是增根∴方程无解

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/(x-1)]^2-2[x/(1-x)]-3=0[x/(x-1)]^2+2[x/(x-1)]-3=0设t=x/(x-1),则x=t/(t-1)t^2+2t-3=0(t-1)(t+3)=0tS

选Dy+3/y-2=0

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

可设x-1为y,那么x=y+1.带入原方程中得：(y/y+1)-{3(y+1)/y+1}=0;简化得：-(y²+5y+3)/y²+y=0;分母y²+y≠0,所以,y≠-1

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)

用换元法设x*x+11x-8＝A然后再做

答：x²-3x-3/x+1/x²+4=0两边同乘以x²得：x^4-3x³+4x²-3x+1=0(x-1)(x³-2x²+2x-1)

(1-3x)2+(2x-1)(1+2x)-5x=01-6x+9x²+4x²-1-5x=013x²-11x=0x(13x-11)=0x=0,x=11/13

方程-1/3+x=2x!

x=-1/3再问：解释再答：

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平方+1/x平方-3/x-3x+4=0x²+2+1/x²-3（x+1/x）+2=0(x+1/x)²-3(x+1/x)+2=0(x+1/x-1)(x+1/x-2)=0x+

（1）原方程即为：(x2-1)/(-2x)+(x+1)/(2x-1)=0即为：(x2-1)/(2x)=(x+1)/(2x-1)即：(x+1)(x-1)(2x-1)=(2x)(x+1)双方除以（x+1）