若(m-1)^3+2m-4=0,(n-1)^3+2n=0.求m+n的值
发布网友
发布时间：2024-02-28 06:27
共2个回答
热心网友
时间：2024-12-03 05:09
(m-1)^3+2(m-1)-2=0 ......(1)
(n-1)^3+2n=0
(n-1)^3+2(n-1)+2=0 ......(2)
(1)+(2)得：
(m-1)^3+(n-1)^3+2(m-1)+2(n-1)=0
{(m-1) + (n-1)} { (m-1)^2 -(m-1)(n-1) +(n-1)^2 } + 2(m+n-2) = 0
(m+n-2)(m^2-2m+1 - mn+m+n-1 +n^2-2n+1) + 2(m+n-2) = 0
(m+n-2)(m^2-2m+1 - mn+m+n-1 +n^2-2n+1+2) = 0
(m+n-2)(m^2-mn+n^2-m-n+3) = 0
(m+n-2) { (1/4m^2-mn+1/4n^2) +3/4(m^2-4/3m)+3/4(n^2-4/3n) + 3 } = 0
(m+n-2) { (1/2m-1/2n)^2 +3/4(m^2-4/3m+4/9)-1/3+3/4(n^2-4/3n+4/9) -1/3+ 3 } = 0
(m+n-2) { (1/2m-1/2n)^2 +3/4(m-2/3)^2)+3/4(n-2/3) ^2 + 7/3 } = 0
(1/2m-1/2n)^2 +3/4(m-2/3)^2)+3/4(n-2/3) ^2 + 7/3 ＞0
∴m+n-2 = 0
m+n = 2
热心网友
时间：2024-12-03 05:10
(n-1)^3+2n=0
2式相加，得
(m-1)^3+2m-4+(n-1)^3+2n=0
(m-1)^3+2m-2+(n-1)^3+2n-2=0
(m-1)^3+2(m-1)+(n-1)^3+2(n-1)=0
(m-1)[(m-1)^2+2]+(n-1)[(n-1)^2+2]=0
∵[(m-1)^2+2]，[(n-1)^2+2]不能等于0
∴m=1，n=1
m+n=2
