设m=5分之1n,求m+2n分之2n+2n-m分之m+4n^2-m^2分之4mn

Question

设m=5分之1n,求m+2n分之2n+2n-m分之m+4n^2-m^2分之4mn

Answer 1

2n/(m+2n)+m/(2n--m)+4mn/(4n^2--m^2)
=2n/(m+2n)--m/(m--2n)--4mn/(m+2n)(m--2n)
=[2n(m--2n)--m(m+2n)--4mn]/(m+2n)(m--2n)
=[2mn--4n^2--m^2--2mn--4mn]/(m+2n)(m--2n)
=--(m^2+4mn+4n^2)/(m+2n)(m--2n)
=--(m+2n)^2/(m+2n)(m--2n)
=--(m+2n)/(m--2n)
当m=n/5时,原式的值＝--[n/5+2n]/[n/5--2n]
=--(n+10n)/(n--10n)
=--(11n)/(--9n)
=11/9.