已知m\n=5\3,求(m\m+n)+(m\m-n)-(n@\m@-n@)的值,@代表平方.

问题描述:

已知m\n=5\3,求(m\m+n)+(m\m-n)-(n@\m@-n@)的值,@代表平方.


m/n=5/3 得 m=5n/3
(m/m+n)+(m/m-n)-(n²/m²-n²)
=[m(m-n)+m(m+n)-n²]/(m²-n²)
=(m²-mn+m²+mn-n²)/(m²-n²)
=(2m²-n²)/(m²-n²)
=[(2*25n²/9)-n²]/[25n²/9-n²]
=41n²/(16n²/9)
=939/16
应该是这样吧

原式
=(m(m-n)+m(m+n)-n^2)/(m^2-n^2)
=(2m^2-n^2)/(m^2-n^2)
=(2(m/n)^2-1)/((m/n)^2-1)
=(2(25/9)-1)/((25/9)-1)
=41/16

(m\m+n)+(m\m-n)-(n@\m@-n@)=m(m-n)/(m+n)(m-n)+m(m+n)/(m+n)(m-n)-n^2/(m+n)(m-n)=(m^2-mn+m^2+mn-n^2)/(m+n)(m-n)=(2m^2-n^2)/(m^2-n^2) m\n=5\3 so 3n=5m 设n=5k 那么 m=3k(2m^2-n^2)/(m^2-n^2) =(2*9k^...