当 k= 时,分式方程 xx−1 kx−1 −xx 1 =0 有增根 .
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/15 03:41:45
方程去分母得,k+2(x-3)=4-x,解得x=10−k3,当分母x-3=0即x=3时方程无解,所以10−k3=3时方程无解,解得k=1.
(1-k)x=4-m当1-k≠0,即k≠1,m为任何数时,方程有唯一解;当1-k=0且4-m=0即k=1且m=4时,方程有无数解;当1-k=0且4-m≠0即k=1且m≠4时,方程无解.
∵分式方程无解,∴x=0或x=1.原方程可化为6x=x+k-3(x-1),整理得,k=8x-3.当x=0时,k=8x-3=-3;当x=1时,k=8-3=5.
K=1两边同乘以x^2-4,得2(x+2)+kx=3(x-2)即kx=x-10故k=1时,方程无解
由题意得x(x-1)≠0,解得x≠0或x≠1.
kx+m=(2k-1)x+4先化简(k-(2k-1))x=4-m(1-k)x=4-mk≠1,方程有唯一解k=1,m=4,方程有无数个解k=1,m≠4,方程无解
两边乘x-2k-2=k(x-2)+1k-2=kx-2k+1kx=3k-3k=0时,是0=0-3,不成立,无解k≠0,x=(3k-3)/k,无解则这是增根,即分母为0x-2=0x=2所以(3k-3)/k
x2+2kx+(k-1)2=0有实数根△=4k²-4(k-1)²≥0k²-(k-1)²≥02k-1≥0k≥1/2
方程两边都乘以x(x+1)得,x2-(x+1)2=k,∵分式方程有增根,∴x(x+1)=0,解得x=0或x=-1,x=0时,k=0-1=-1,x=1时,k=1-0=1,所以,k=±1.
两边乘x-2k-2=k(x-2)+1k-2=kx-2k+1kx=3k-3k=0时,是0=0-3,不成立,无解k≠0,x=(3k-3)/k,无解则这是增根,即分母为0x-2=0x=2所以(3k-3)/k
方程两边都乘以(x-1)(x+1),得x(x+1)+k(x+1)-x(x-1)=0.解得x=-kk+2,∵分式方程无解,∴−kk+2=±1,解得k=-1,故答案为:-1.
kx²-K(x+2)=x(x+1)+6kx²-Kx-2k=x²+x+6kx²-x²-kx-x-2k-6=0(k-1)x²-(k+1)x-2k
两边乘x(x+1)3(x+1)+x(kx+3)=2x(x+1)有增根则分母等于0x(x+1)=0x=0,x=-1x=0代入3(x+1)+x(kx+3)=2x(x+1)3=0,不成立x=-1代入3(x+
设x1,x2为方程两根x1>2x2>2则(x1-2)(x2-2)>0x1+x2>4x1x2-2(x1+x2)+4>0x1+x2=2kx1x2=k^2-12k>4k>2x1x2-2(x1+x2)+4=k
两边乘(x+1)(x-1)x(x+1)+k(x+1)-x(x-1)=02x+kx+k=0无解则是增根所以分母为0x=-1或1x=-1-2-k+k=0,不成立x=12+k+k=0k=-1
分式xx−1有意义,则x-1≠0,解得:x≠1,故答案为:≠1.
K不等于-1或2的时候
方程两边都乘以(x-1)(x+2)得,x(x+2)-(x-1)(x+2)=m,x2+2x-x2-x+2=m,m=x+2,∵分式方程有增根,∴(x-1)(x+2)=0,∴x-1=0,x+2=0,解得x1
分式方程2+1-kx/x-2=1/2-x有增根∴增根是x=2原方程去分母得2(x-2)+1-kx=-1x=2代入得0+1-2k=-1k=1
是这个方程?2/(x-2)+(kx)/(x^2-4)=3/(x+2)?是问方程无解吗?两边都乘以x^2-4,得2(x+2)+kx=3(x-2)即kx=x-10所以当k=1或k=-4(x=2)或k=6(