解方程x的平方-1分之x-x分之3x的平方-3-搜答案-灵鹊做题网

用解比例的方法与平方根的方法化成x*x/x+2=x+1/2xx*x+2x=2x+3方程两边同时减去2xx*x=3x=3/xx=0够详细了吧

1、令a=x+1/xa²=x²+2+1/x²x²+1/x²=a²-2所以a²-2+a=0(a+2)(a-1)=0a=-2,a=1x

6/(x+1)=(x+5)/x(x+1);(6x-x-5)/x(x+1)=0;(5x-5)/x(x+1)=0;∴x=1;很高兴为您解答,skyhunter002为您答疑解惑如果本题有什么不明白可以追问

3/x+3/(x-1)=(x+5)/(x^2-x)[3(x-1)+3x]/[x(x-1)]=(x+5)/[x(x-1)]当x(x-1)≠0,即x≠0,且x≠1时,去分母可得3(x-1)+3x=x+53

(xx-8x+16)/(xx-4x+4)+[1+2/(x-2)][1+2/(x-2)]=2x/(x-2)(x-4)^2/(x-2)^2+[1+2/(x-2)]^2=2x/(x-2)[1-2/(x-2)

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

1-x分之3-x+x的平方-8x+7分之x的平方-2=1+7-x分之5-x(x-3)/(x-1)+(x²-2)/[(x-7)(x-1)]=1+(x-5)/(x-7)平方同乘以(x-1)(x-

6/(x-1)+3/x=(x+5)/(x²-x)方程式两边都乘以x(x-1),分母不能等于0,即x≠0且x≠1得：6x+3(x-1)=x+5化简得：6x+3x-3=x+5,即8x=8得：x=

1:x²+X-2分之3=X-1分之X-X+2分之XX²+X-(X-1分之X)+X-2分之X=0X²+2X-(X-1分之X)-2分之X=0∵X-1≠0所以（X-1)(X&#

两边乘以x(x-1)得3x-(x+2)=03x-x-2=02x=2∴x=1检验：x=1是增根∴原方程无解

解方程：(1)(x²-5x)/(x+1)+24(x+1)/x(x-5)+14=0；(2).2(x²+1)/(x+1)+6(x+1)/(x²+1)=7；(3)(x̾

x的平方-1分之4-x-1分之x+1=-14-(x+1)(x+1)=-(X+1)(X-1)4-x^2-2x-1=-x^2+12x=2x=1经检验x=1是方程的增根

有理化：（这种叙述可以有多种歧义,我就默认是这种吧![x/(x-2)]=[2x/(x-3)]+[(1-x²)/(x²-5x+6)])各项乘以（x-2)（x-3):x(x-3)=2x

因为x-x=0,所以原式=0-2x/6=12x/6=-12x=-6x=-3

5x²-4x-1/4=x²+3x/4即16x²-19x-1=0即x=(19±√17)/32

(x+1)/（x-1）+4/（x²-1）=1（x²+2x+1+4）/（x²-1）=1x²+2x+5=x²-12x=-6x=-3

两边乘以(x+2)(x-2)得-x²=(x-2)-(x+2)(x-2)-x²=x-2-x²+4x=-2检验：x=-2是增根∴方程无解再问：��ʽa-3��֮

x²+3/(2x)-2x²-1/(2x)=0-x²+1/x=0x²=1/xx³=1x=1

两边乘以(x+2)(x-2)得x(x-2)-(x-2)²=kx²-2x-x²+4x-4=k2x=k+4∵无解∴x=-2或2当x=-2时k=-8当x=2时k=0