x平方-1除以x-1 1是什么间断点-搜答案-智慧作业帮

数学之美团为你解答x/(x-2)-1=8/(x²-4)同乘以x²-4x(x+4)-(x²-4)=8x²+4x-x²+4=84x=4x=1经检验,x=1

x+2分之x平方-2x-x-1分之x平方-4X+4除以x平方-16分之x平方+4x=x(x-2)/(x+2)÷(x-2)^2/(x-1)÷[x(x+4)/(x+4)(x-4)]=[x(x-2)/(x+

5/(x²+x)-1/(x²-x)=05/x(x+1)-1/x(x-1)=05x-5-x-1=04x=6x=3/2x=2分之3再问：5x-5-x-1=0这个是怎么算的？再答：方程两

((x+2)/(x^2-2x)-(x-1)/(x^2-4x+4))/((x^2-16)/(x^2+4x))=((x+2)(x-2)-(x-1)x)/((x-2)^2*x)/((x-4)/x)=((x-

二分之根号2乘以arctan[(x-1)/根号(2x)]+四分之根号2乘以lnabs[(x+根号2x+1)/(x-2x+1)]+C

格式有点乱,〖((1-x)/(3-x))^2/((x^2-6x+9)/(9-x^2 ))〗^ ×1/(x^2-2x+1)=〖((1-x)/(3-x))^2/(〖(X-3)〗^2/(

4x三次除以(-2x)平方-(2x平方-x)除以(1/2x)=4x^3/4x^2-(2x^2-x)/(1/2x)=x-2x(2x-1)/x=x-2(2x-1)=x-4x+2=-3x+2

1除以(x的平方+3x+2)+1除以(x平方+5x+6)+1除以(x平方+7x+12)=1/（x+1）（x+2）+1/(x+2)(x+3)+1/(x+3)(x+4)=1/(x+1)-1/(x+2)+1

题目是（5x+2)/(x²+x)=3/(x+1)这样吧?∵（x+1）作分母,不为0,可直接约分∴（5x+2)/x=3∴5x+2=3x∴2x=-2∴x=-1

=[(x+1)(x-1)/(x-1)²-(x-1)/(x+1)]×(x-1)/x=[(x+1)²-(x-1)²]/(x+1)(x-1)×(x-1)/x=4x/(x+1)(

先说答案x1=2x2=1/22(x*2+1/x*2)-3(x+1/x)-1=02(x+1/x)*2-4-3(x+1/x)-1=0(x+1/x)*2-3/2(x+1/x)-5/2=0(x+1/x-3/4

(x²-6x+9)/(x+1)÷(x²-9)/(x²+x)=[(x-3)²/(x+1)]×[x(x+1)/(x+3)(x-3)]=x(x-3)/(x+3)x的平

将3x/(x-1)-x/(x+1)通分得

x²/(x-1)-x=(x²-x²+x)/(x-1)=x/(x-1)

2x分之x的平方-4x+4除以x平方分之x的平方-2x+1等于2分之x,当x=0或2时,分式无意义

[(x²-4x+4)/(2x)]÷[(x²-2x+1)/x²]=[(x-2)²/(2x)]/[(x-1)²/x²]=(x-2)²x

y=(3x²+2x+2)/(x²+x+1)yx²+yx+y=3x²+2x+2(y-3)x²+(y-2)x+(y-2)=0x是实数,所以判别式大于等于0

x的平方-1分之x除以x的平方分之x的平方=x²-(x/1)/(x²/x²）=x²-(x)/1=x²-x