25000(1-x)^2=0 解

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

1,（1）将（x+3/x-1）设为a（2）得a²-5a+6=0（3）因式分解（a-2）（a-3）=0（4）a1=2a2=32,化简得x³-4x²+18=0

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~~

①0=x(x+1)(x+2)(x+3)-3=x(x+3)(x+1)(x+2)-3=(x^2+3x)(x^2+3x+2)-3=(x^2+3x)^2+2(x^3+3x)-3=(x^2+3x+3)(x^2+

(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

1.方程两边同时乘(2x-1)(x-2)得2x(x-2)+x(2x-1)=2(x-2)(2x-1)解得x=4/5检验x=4/5时(2x-1)(x-2)≠0∴x=4/5为原分式方程的解2.方程两边同时乘

x/(1-x)+x/(1+x)+2x/(1+x^2)+4x/(1+x^4)=0[x(1+x)+x(1-x)]/(1-x^2)+2x/(1+x^2)+4x/(1+x^4)=02x/(1-x^2)+2x/

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)

有种方法叫做穿针引线法,需要在纸上画出函数的简图,这个题的答案应该是-2,3,6,1

x²+x-12=0a=1,b=1,c=-12△=b²-4ac=1²-4×1×(-12)=1+48=49√△=√49=7x=(-b±√△)/(2a)=(-1±7)/2x1=

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

(x+6)/(x+1)-(3x²+10x+4)/(x²+3x+2)+(2x+1)/(x+2)=0(x+6)/(x+1)-(3x²+10x+4)/(x+1)(x+2)+(2

=x^2002(x^4+x^3+x^2+x+1)=0提取公因式就行了

3x²=10x+83x²-10x-8=0x=(10±√(100+4×3×8)/2*3x=(10±√196)/6x=(10±14)/6∴x1=4x2=-2/33x²-10x