若使分式(x-2) (x^2-2x c)
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X=-6x=-6(X/(X-2))*((X-6)/(X-3))=1-
原分式方程化为:2x+1x(x+1)=56(x+1),方程的两边同乘6x(x+1),得6(2x+1)=5x,解得x=-67.检验:把x=-67代入6x(x+1)≠0,即x=-67是原分式方程的解.则原
去分母得:2+x(x+2)=x2-4,解得:x=-3,经检验x=-3是分式方程的解.
原式=[x²/(x-1)]-x-1=[x²/(x-1)]-[(x+1)(x-1)]/(x-1)=[x²/(x-1)]-(x²-1)/(x-1)=[x²
当分母x-6≠0,即x≠6时,分式2x−6有意义;故答案是:x≠6.
x²/(x+2)-x+2=x²/(x+2)-(x-2)=[x²-(x+2)(x-2)]/(x+2)=[x²-(x²-4)]/(x+2)=4/(x+2)
(x²-4)/(x+1)(x-2)]=0[(x+2)(x-2)]/[(x+1)(x-2)]=0(x+2)/(x+1)=0∴x+2=0x=-2再问:这是初二数学。第一课分式。。请您在清楚点。。
∵当△=b2-4ac<0时,x2+4x+m=0无解,即42-4m<0,解得m>4,∴当m>4时,不论x取何实数,分式总有意义.故答案为m>4.
分式x+5/(x+1)(x+2)有意义,则分母不为0,即x≠-1且x≠-2x的绝对值+2/(x-2)(x+3)值为零,不可能吧5-x/x^2值为正数,则5>x/x^2,分母x^2大于0(不能等于0),
(X/(X-2))*((X-6)/(X-3))=1->X(X-6)=(X-2)(X-3)->X^2-6X=X^2-5X+6->X=-6
不是x≠2或x≠-1应该是:x≠2且x≠-1
分式2x/x^2-1除以2x/1-x分式有意义,所以必须满足x≠0分式x(x-2)/x^2+2的值为0x(x-2)=0x=0(设)或x=2所以2x/x^2-1除以2x/1-x=(4/3)除以(-4)=
∵2x−1x+1=2-3x+1,∴根据题意,得x+1=±1,±3,解得x=0,-2,2,-4,又∵x取整数,∴x的整数值为0,-2,2,-4.再问:лл��
(|x|-2)/(x²-x-2)=0|x|-2=0x=±2x²-x-2≠0x≠2且x≠-1综上x=-2
1.A.(X^2)/X=X或-XB.(X-2)/(X+1),是最简分式C.X/(X-1)是最简分式D.(2X+2)/(4X)=(X+1)/(2X)你是不是写错了?B和C都是最简分式2.A.(M^2-N
∵2x+2x2−1=2x−1,∴根据题意,得x-1=±1或±2,则x=2或0或3或-1.又x≠±1,则x=0或2或3.
∵分式2x+12x−1无意义,∴2x-1=0,∴x=12.故答案为:12.
原式=(X+1)(X-1)/(X+1)^2=(X-1)/(X+1)
2xy/x+y2x.2y/2x+2y=4xy/2*(x+y)=2xy/x+y不知道对不对
【参考答案】原式=[(x^2+x)/(x^2-1]-[x^2/(x^2-2x+1)]=[(x^2+x)(x-1)/(x^2-1)(x-1)]-[x^2(x+1)/(x^2-2x+1)(x+1)]=[(