已知㎡=n+2,n²=m+2﹙m≠n﹚,求∶m+n; m³-2mn+n³

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已知㎡=n+2,n²=m+2﹙m≠n﹚,求∶m+n; m³-2mn+n³

已知㎡=n+2,n²=m+2﹙m≠n﹚,求∶m+n; m³-2mn+n³
已知㎡=n+2,n²=m+2﹙m≠n﹚,求∶m+n; m³-2mn+n³

已知㎡=n+2,n²=m+2﹙m≠n﹚,求∶m+n; m³-2mn+n³
已知㎡=n+2,n²=m+2﹙m≠n﹚,那么:
m²-n²=n-m
即(m-n)(m+n)=-(m-n)
由于m≠n,所以可得:m+n=-1
那么:m²+n²=n+2+m+2=m+n+4=3
而2mn=(m+n)²-(m²+n²)=1-3=-2
则得:mn=-1
所以:m³-2mn+n³
=(m+n)(m²-mn+n²)-2mn
=-(3-mn)-2mn
=-3-mn
=-3+1
=-2


m²=n+2 (1)
n²=m+2 (2)
(1)-(2)
m²-n²=n-m
(m+n)(m-n)+(m-n)=0
(m-n)(m+n+1)=0
m≠n,m-n≠0,要等式成立,只有m+n+1=0
m+n=-1

(1)×(2)
m²n²...

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m²=n+2 (1)
n²=m+2 (2)
(1)-(2)
m²-n²=n-m
(m+n)(m-n)+(m-n)=0
(m-n)(m+n+1)=0
m≠n,m-n≠0,要等式成立,只有m+n+1=0
m+n=-1

(1)×(2)
m²n²=(n+2)(m+2)=mn+2(m+n)+4=mn+2(-1)+4=mn+2
m²n²-mn-2=0
(mn-2)(mn+1)=0
mn=2或mn=-1
mn=2时,m,n是方程x²-x+2=0的两根,方程无实根。
mn=-1

m³-2mn+n³
=(m+n)(m²-mn+n²) -2mn
=(m+n)[(m+n)²-3mn]-2mn
=(-1)[(-1)²-3mn]-2mn
=3mn-1-2mn
=mn
=-1

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m" = n + 2
n" = m + 2
那么
m" - n" = (n + 2) - (m + 2) = n - m
还有
m" - n" = (m + n)(m - n)
n - m = -(m + n)(n - m)
因为 m ≠ n,那么
m + n = -1

再看
m" = n + 2
...

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m" = n + 2
n" = m + 2
那么
m" - n" = (n + 2) - (m + 2) = n - m
还有
m" - n" = (m + n)(m - n)
n - m = -(m + n)(n - m)
因为 m ≠ n,那么
m + n = -1

再看
m" = n + 2
n" = m + 2

m" - n = 2
n" - m = 2
那么
m^3 - 2mn + n^3
= m^3 - mn + n^3 - mn
= m(m" - n) + n(n" - m)
= 2m + 2n
= 2(m + n)
= 2 X (-1)
= -2

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