方程x^2 2x-1=0-搜答案-智慧作业帮

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

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~~

只有一个实根.设f(x)=x^3+2x-19为单调增函数.所以只有一个实根.下面来求这个实根由于f(2)=-7f(3)=11所以这个根在（2,3）内.利用二分法求这个解.取x0=5/2f(5/2)=1

选Dy+3/y-2=0

即x/(2x-5)-5/(2x-5)-1=0两边乘2x-5x-5-(2x-5)=0x=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)

x²+39=(x+1)²-22x²+39=x²+2x+1-222x=60x=30

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

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/x-1)-(2x-2/x)-1=0（x/x-1)-2(x-1)/x-1=0设（x/x-1)=y则y-2/y-1=0两边乘yy^2-y-2=0(y-2)(y+1)=0y=2或y=-1则x/x-1=

x(x+1)(x²-2x-4)=0x1=0x2=-1x²-2x-4=0x²-2x+1=5(x-1)²=5x3=1+√5x4=1-√5x³-2x+1=0

x(x+1)-x-9=0x²+x-x-9=0x²-9=0(x-3)(x+3)=0x=3或x=-3x²-2x=224x²-2x-224=0(x-16)(x+14)

用换元法设x*x+11x-8＝A然后再做

方程：x+1/8x=22

再问：真的很谢谢，懂了

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^2-x-1=0a=1,b=-1,c=-1b^2-4ac=1+4=5所以x=(1±√5)/25x^2-8x+2=0a=5,b=-8,c=2b^2-4ac=64-40=24所以x=(8±√24)/10

（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）