已知:a,b满足a^3-3a^2+5a=1,b^3-3b^2+5b=5,则a+b=____.

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已知:a,b满足a^3-3a^2+5a=1,b^3-3b^2+5b=5,则a+b=____.

已知:a,b满足a^3-3a^2+5a=1,b^3-3b^2+5b=5,则a+b=____.
已知:a,b满足a^3-3a^2+5a=1,b^3-3b^2+5b=5,则a+b=____.

已知:a,b满足a^3-3a^2+5a=1,b^3-3b^2+5b=5,则a+b=____.
x^3-3x^2+5x=(x-1)³+2(x-1)+3
(a-1)³+2(a-1)+3=1
(b-1)³+2(b-1)+3=5
相加得到
(a-1)³+2(a-1)+(b-1)³+2(b-1)=0
(a+b-2)[(a-1)²-(a-1)(b-1)+(b-1)²]+2(a+b-2)=0
(a+b-2)[(a-1)²-(a-1)(b-1)+(b-1)²+2]=0
[(a-1)²-(a-1)(b-1)+(b-1)²]+>0
故a+b=2

∵ x³-3x²+5x=(x-1)³+2(x-1)+3
∴ (a-1)³+2(a-1)+3=1 …… ①
(b-1)³+2(b-1)+3=5 …… ②
①+②得:令a-1 = m ; b-1 = n
m³+2m+n³+2n=0
(m+n)[m²-mn+n²]+...

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∵ x³-3x²+5x=(x-1)³+2(x-1)+3
∴ (a-1)³+2(a-1)+3=1 …… ①
(b-1)³+2(b-1)+3=5 …… ②
①+②得:令a-1 = m ; b-1 = n
m³+2m+n³+2n=0
(m+n)[m²-mn+n²]+2(m+n)=0
(m+n)[m²-mn+n²+2]=0
∵m²-mn+n² > 0 恒成立
∴ m+n=0
即:a-1+b-1= 0
∴ a+b = 2

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