(m除以m+1)+(2除以m-1)=1怎么做

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(m除以m+1)+(2除以m-1)=1怎么做
(m除以m+1)+(2除以m-1)=1怎么做

(m除以m+1)+(2除以m-1)=1怎么做
m/(m+1)+2/(m-1)=1
两边同乘以(m+1)(m-1)
m(m-1)+2(m+1)=m²-1
m²-m+2m+2=m²-1
m=-3
检验符合

m/(m+1)+2/(m-1)=1
两边都乘以(m+1)(m-1)=>
m(m-1)+2(m+1)=m^2-1
=>m=-3

m/(m+1)+2/(m-1)=1
两边同乘以(m+1)(m-1)
m(m-1)+2(m+1)=m^2-1
m^2+m+2-m^2+1=0
m=-3

m/(m+1) + 2/(m-1) = 1
[m(m-1)+2(m+1)]/(m+1)(m-1)=1
m²+m+2=m²-1
m=-3

m/(m+1)+2/(m-1)=1
m(m-1)/((m+1)(m-1))+(m+1)/((m+1)(m-1))=1
(m²-m+2m+2)/(m²-1)=1
m²+m+2=m²-1
所以m=-3

原式=[m(m-1)+2(m+1)]/m*m-1=1
即m*m+m+2=m*m-1
m=-3