2个回答
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设原式=t
所以:0<t≤√2
则:t=根号[2-根号(2+t)]
t²=2-根号(2+t)
(t²-2)²=2+t
t^4-4t²+4=2+t
t^4-4t²-t+2=0
(t^4-4t²+3)-(t+1)=0
(t²-3)(t²-1)-(t+1)=0
(t²-3)(t-1)(t+1)-(t+1)=0
(t+1)(t³-t²-3t+2)=0
(t+1)(t³-2t²+t²-3t+2)=0
(t+1)[t²(t-2)+(t-2)(t-1)]=0
(t+1)(t-2)(t²+t-1)=0
t+1=0或t-2=0或t²+t-1=0
t=-1(舍去)
t=2 (舍去)
t²+t-1=0
(t+1/2)²=5/4
t+1/2=√5/2
t=(√5-1)/2
所以:原式=(√5-1)/2
所以:0<t≤√2
则:t=根号[2-根号(2+t)]
t²=2-根号(2+t)
(t²-2)²=2+t
t^4-4t²+4=2+t
t^4-4t²-t+2=0
(t^4-4t²+3)-(t+1)=0
(t²-3)(t²-1)-(t+1)=0
(t²-3)(t-1)(t+1)-(t+1)=0
(t+1)(t³-t²-3t+2)=0
(t+1)(t³-2t²+t²-3t+2)=0
(t+1)[t²(t-2)+(t-2)(t-1)]=0
(t+1)(t-2)(t²+t-1)=0
t+1=0或t-2=0或t²+t-1=0
t=-1(舍去)
t=2 (舍去)
t²+t-1=0
(t+1/2)²=5/4
t+1/2=√5/2
t=(√5-1)/2
所以:原式=(√5-1)/2
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