(1+n/m-n/m-n)÷(1-n/m-m/m+n) 化简,结果

2个回答

  • =(1+n/m-n)×(1/m-n)除以(1-N/m-n)×(1/m+n)=[1+N/(m-n)²]除以[(1-N)/(m²-n²)]=[(1+n)/(m-n)²]×[(m²-n²)/(1-n)=(1+n)/(m-n)²]×﹛[(m-n)²/(1-n)]+[(2mn-2n²)/(1-n)]﹜=(1+n)/(1-n)+[(2mn-2n²)/(1-n)]×[(1+n)/(m-n)²]=(1+n)/(1-n)+[2n(m-n)/(1-n)]×[(1+n)/(m-n)²]=(1+n)/(1-n)+[2n(1+n)]/[(1-n)×(m-n)]=[(2n+1)×(1+n)]/[(1-n)×(m-n)],水平有限,只能做到这,求高手.