Exercise:
Solve in R , (x2−5x+4)2−3(x2−5x+4)+1=x
Solution: Let u(x)=x2−5x+4
So u(x)=x2−4x−x+4=x(x−1)−4(x−1)=(x−1)(x−4)
Thus (x−1)2(x−4)2−3(x−4)(x−1)−(x−1)=0
⇔(x−1)[(x−1)(x−4)2−3(x−4)−1]=0
⇔x=1or(x−1)(x−4)2−3(x−4)−1=0
⇔x(x−4)2−(x−4)2−3(x−4)−1=0⇔(x−4)(x(x−4)−3)−(x−4)2−1=0
⇔(x−4)(x2−4x−3)−(x−4)2−1=0⇔(x−4)[x2−4x−3−x+4]−1=0
⇔(x−4)(x2−5x+1)−1=0⇔(x−4)(x(x−5)+1)−1=0
⇔x(x−5)(x−4)+x−4−1=0⇔(x−5)(x(x−4)+1)=0
So x=5orx2−4x+4−4+1=0⇔(x−2)2−3=0⇔x=2±√3
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