用换元法解方程2x÷x-1
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/18 12:57:42
原方程等价于2x+8/x-5=0因为x不为零,可以两边同时乘以x,得到2x²-5x+8=0这个方程无解所以原方程无解原方程无解是肯定无疑的,如果楼主学过均值不等式或者导数,就可以知道:f(x
令t=x/(x²-1);则有:3t+1/(2t)=5/2;6t²+1-5t=0;(2t-1)(3t-1)=0;t=1/2或t=1/3;∴x/x²-1=1/2;x²
设3x^+2x+1=a则a-2=3/a解得a=3或-1∴3x^+2x+1=3或-1当3x^+2x+1=3得x=(-1±√7)/3当3x^+2x+1=-1得无解∴x=(-1±√7)/3
设3x/(x^2-1)=t,则t+1/t=5/2,去分母得t^2-5t/2+1=0,即(t-2)(t-1/2)=0,所以t=2或t=1/2,当t=2时得3x/(x^2-1)=2,即2x^2-3x-2=
X^2+1/x^2-2(x+1/x)-1=0(X+1/X)^2-2(X+1/X)-3=0x+1/x=YY^2-2Y-3=0(Y-3)(Y+1)=01)Y=3X+1/X-3=0X^2-3X+1=0X=(
令(x+1)/x^2=y则x^2/(x+1)=1/y原方程可化为y-2/y=1即y^2-y-2=0(y-2)(y+1)=0y=2或y=-1当y=2即(x+1)/x^2=2则2x^2-x-1=0(2x+
设3x平方/x+1=T则方程为T-(3/T)=2同分母得T平方-2T-3=0所以T=-1或3若T=3则3x平方/x+1为3这种有分数的方程要同分母才能求值x=.(好像有根号)若T=-1则3x平方/x+
设x2+2x=y,则原方程为y-6y=1.解之得,y1=3,y2=-2.则x2+2x=3或x2+2x=-2,当x2+2x=3时,有x2-3x+2=0,解之得,x1=1,x2=2.当x2+2x=−2时,
令a=x²+x则3a=2/a+13a²-a-2=0(3a+1)(a-1)=0a=-1/3,a=1x²+x=-1/33x²+3x+1=0无解x²+x=1
设y=x2+x,则得y+1=2y,方程两边同乘以y,整理得y2+y-2=0.故本题答案为:y2+y-2=0.
[x/(x-1)]^2-2[x/(1-x)]-3=0[x/(x-1)]^2+2[x/(x-1)]-3=0设t=x/(x-1),则x=t/(t-1)t^2+2t-3=0(t-1)(t+3)=0tS
(2x÷x²-1)-1÷x²-x=0(2/x-1)-1/x²-x=02/x-1-1/x²-x=0x+1-2/x+1/x²=0x³+x
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)
由2x−1x=y可得x2x−1=1y.所以原方程可化为y-1y=2,整理得y2-2y-1=0.
令x-2/x=t,则t^2-2t+1=0,t=1,所以x-2/x=1,即x^2-x-2=0,解得x=2,或x=-1
设x−1x+3=y,那么x+3x−1=1y,原方程变形为2y+6y=7,整理得2y2-7y+6=0.解这个方程,得y1=32,y2=2.当y=32时,x−1x+3=32,去分母,得3x+9=2x-2,
设x-2/x=t则t-3/t=2t^2-3=2tt^2-2t-3=0(t-3)(t+1)=0t=3或-1若t=3x-2/x=3,x=-1若t=-1x-2/x=-1,x=1经检验,x1=-1x2=1均为
设y=x²+2x,则原方程可变为:y-6/y=1y²-6=yy²-y-6=0(y+2)(y-3)=0y=-2y=3所以x²+2x=-2解这个方程无解x²
令a=x²-1则x²-2=a-1所以a-1=2/a两边乘aa²-a-2=0(a+1)(a-2)=0a=-1,a=2x²-1=-1x²=0x=0x
=[3/(x+1)-(x-1)]÷(x²-2x)/(x+1)=[3/(x+1)-(x-1)(x+1)/(x+1)]÷x(x-2)/(x+1)=[3-(x+1)(x-1)]/(x+1)÷x(x