Question

高分求解

Answer 1

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]+2>0
故a+b=2

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两个式子可变形为：
(a-1)^3+(a-1)=-2;
(b-1)^3+(b-1)=2
令f(x)=x^3+x,
可得f(a-1)=-f(b-1),
即a-1=-(b-1),
故a+b=2

Answer 2

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]+2>0
故a+b=2

Answer 3

路过，好像我不会，蛋疼啊

Answer 4

一元三次方程可以解的====================