Question

解方程:2(x^2+1/x^2)-9（x+1/x)+14=0(提示：令x+1/x=t)

Answer 1

2(x^2+1/x^2)-9（x+1/x)+14=0
令x+1/x=t
∴x^2+1/x^2=t²-2
原方程为
2(t²-2)-9t+14=0
2t²-4-9t+14=0
2t²-9t+10=0
(2t-5)(t-2)=0
2t=5=0 t-2=0
t=5/2 或t=2
∴x+1/x=5/2或 x+1/x=2
2x²-5x+2=0 x²-2x+1=0
(2x-1)(x-2)=0 (x-1)²=0
2x-1=0 x-2=0 x-1=0
∴x₁=1
x₂=2
x₃=1/2

追问

2x²-5x+2=0 x²-2x+1=0
(2x-1)(x-2)=0 (x-1)²=0
2x-1=0 x-2=0 x-1=0

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Answer 2

2(x²+2+1/x²)-9(x+1/x)+10=0
2(x+1/x)²-9(x+1/x)+10=0
令x+1/x=t
∴原方程可化为
2t²-9t+10=0
(2t-5)(t-2)=0
∴t=5/2 t=2
∴x+1/x=5/2 x+1/x=2
∴x=2 x=1/2 x=1

Answer 3

2（t^2-2）-9*t+14=0
2t^2-9t+10=0
(2t-5)*(t-2)=0
t=2 t=5/2
x=1 x=2 x=1/2