1.若a小于b,则a-b的相反数的倒数的绝对值是?2.已知：m^2=n+2,n^2=m+2(m不等于n),求：m^3-2mn+n^3

题目

1.若a小于b,则a-b的相反数的倒数的绝对值是?2.已知：m^2=n+2,n^2=m+2(m不等于n),求：m^3-2mn+n^3

答案

|1/[-(a-b)]|
=|1/(b-a)|
=1/(b-a)
m^2=n+2,n^2=m+2
相减
m^2-n^2=n-m
(m-n)(m+n)+(m-n)=0
(m-n)(m+n+1)=0
因为m不等于n
所以m+n+1=0
m+n=-1
m^3-2mn+n^3
=(m+n)(m^2-mn+n^2)-2mn
=-(m^2-mn+n^2)-2mn
=-m^2-mn-n^2
=-m(m+n)-n^2
=m-n^2
因为n^2=m+2
所以原式=-2