解方程x^2-2x-1=0-搜答案-百度作业网

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

原方程可变为：2x²-18x+16=0即,x²-9x+8=0把方程左边因式分解的,（x-1）（x-8）=0从而的,x-1=0或x-8=0解得,x=1或x=8

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

2x(x-1)-x(3x+2)=-x(x+2)-122x^2-2x-3x^2-2x=-x^2-2x-122x^2-3x^2+x^2-2x-2x+2x=-12-2x+12=0-2(x-6)=0x-6=0

(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

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

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

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^2+x)(x^2+x-3)-3(x^2+x)+8=0（x^2+x)^2-6(x^2+x)+8=0(x^2+x-4)(x^2+x-2)=0x^2+x-4=0x^2+x-2=0(x+1/2)^2=1

(X(13-X)(X+(13-X)))/(X+1)=42得分母是X^2(13-X)+(13-X)X分子是（X+1）再得分母X(13-X)(X+1)分子分母抵消的（X+1）X=42得X=6

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)

你好,我认真解答了这道题,看最终没有实数解啊.我希望能帮到你

（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）

解方程,x(x-2)+x-2=0