用换元法求 ( X/(X-1) )^2-5(X/ (X-1) )^2+6=0 3X/(X^2-1)+(X^2-1)/3X=2/5( X/(X-1) )^2-5(X/ (X-1) )^2+6=0 3X/(X^2-1)+(X^2-1)/3X=2/5

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用换元法求 ( X/(X-1) )^2-5(X/ (X-1) )^2+6=0 3X/(X^2-1)+(X^2-1)/3X=2/5( X/(X-1) )^2-5(X/ (X-1) )^2+6=0 3X/(X^2-1)+(X^2-1)/3X=2/5

用换元法求 ( X/(X-1) )^2-5(X/ (X-1) )^2+6=0 3X/(X^2-1)+(X^2-1)/3X=2/5( X/(X-1) )^2-5(X/ (X-1) )^2+6=0 3X/(X^2-1)+(X^2-1)/3X=2/5
用换元法求 ( X/(X-1) )^2-5(X/ (X-1) )^2+6=0 3X/(X^2-1)+(X^2-1)/3X=2/5
( X/(X-1) )^2-5(X/ (X-1) )^2+6=0
3X/(X^2-1)+(X^2-1)/3X=2/5

用换元法求 ( X/(X-1) )^2-5(X/ (X-1) )^2+6=0 3X/(X^2-1)+(X^2-1)/3X=2/5( X/(X-1) )^2-5(X/ (X-1) )^2+6=0 3X/(X^2-1)+(X^2-1)/3X=2/5
(1) 令t=x/(x-1) 则原方程可化为 t^2 -5t+6=o 解得 t=2 或 t=3
当t=2时,x=2
当t=3时,x=3/2
故x=2或x=3/2
(2)t=3x/(x^2-1) 则原方程可化为 t+1/t=5/2 即t^2 -(5/2)t +1=0 解得t=1/2 或 t=2
当t=1/2时,x= 3加减根号10
当t=2时,x=2 或 - 1/2

1令x/x-1=t,所以
( X/(X-1) )^2-5(X/ (X-1) )^2+6=0,得到t^2-5t+6=0,解得t=2或3,所以
x/x-1=2或3解得x=2或3/2
令3X/(X^2-1)=m,所以
3X/(X^2-1)+(X^2-1)/3X=2/5得到m+1/m=5/2 解得m=1/2或2,所以
3X/(X^2-1)=1/2或2,解得x=3±根号10,x=-1/2或2