1、(x²-x)²-5(x²-x)+6=02、x+1/x² - 2x²/x+1=1
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1、(x²-x)²-5(x²-x)+6=02、x+1/x² - 2x²/x+1=1
1、(x²-x)²-5(x²-x)+6=0
2、x+1/x² - 2x²/x+1=1
1、(x²-x)²-5(x²-x)+6=02、x+1/x² - 2x²/x+1=1
1.设(x²-x)为y,则
y²-5y+6=0
(y+1)(y-6)=0
y1=6 y2= -1
x²-x=6
x²-x-6=0
(x+2)(x-3)=0
x1= -2 x2=3
x²-x= -1
x²-x+1=0
△=b²-4ac=(-1)²-4×1×1= -3<0
所以无解
综合x1 =-2 ,x2 = 3
2.设x+1/x²为y,则
y-2/y=1
同乘y
y²-2=y
y²-y-2=0
(y-2)(y+1)=0
y1 =2 ,y2= -1
x+1/x²=2
x+1/x²-2=0
同乘x²
-2x²+x+1=0
△=b²-4ac=1²-4×(-2)×1=7
后面自己算吧 ,这两道题都是用换元法来算,掌握之后很好用的