∵ 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
如果本题有什么不明白可以追问,
∵ 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
如果本题有什么不明白可以追问,