题目

用换元法解方程(1)x+1分之x平方-5x+x(x-5)分之24(x+1)+14=0(2)x+1分之2(x平方+1）+x平方+1分之6（x+1)=7
(3)x平方分之x4次方+2x+1+x分之x平方+1=2

提问时间：2020-07-24

答案

(1)令(x^2-5x)/(x+1)=t,则(x+1)/(x^2-5x)=1/t,所以原方程换元后为t+24/t+14=0,方程两边同时乘以t得t^2+14t+24=0解得t=-2或t=-12即(x^2-5x)/(x+1)=-2或(x^2-5x)/(x+1)=-12,
解(x^2-5x)/(x+1)=-2得: x=1或x=2
解(x^2-5x)/(x+1)=-12得: x=-3或x=-4
（2）令(x^2+1)/(x+1)=t,(x+1)/(x^2+1)=1/t,所以原方程换元后为2t+6/t=7,方程两边同时乘以t得2t^2+6=7t,解得t=2或t=3/2即(x^2+1)/(x+1)=2或(x^2+1)/(x+1)=3/2
解(x^2+1)/(x+1)=2得: x=1+√2或x=1-√2
解(x^2+1)/(x+1)=3/2得: x=(3+√17)/4或x=(3-√17)/4
(3)(x^4+2x^2+1)/x^2+(x^2+1)/x=(x^2+1)^2/x^2+(x^2+1)/x=[(x^2+1)/x]^2++(x^2+1)/x
令(x^2+1)/x=t,则原方程换元后为t^2+t=2,解得t=-2或t=1即(x^2+1)/x=1或(x^2+1)/x=-2
解(x^2+1)/x=1得：无解
解(x^2+1)/x=-2得：x=-1