`h)x/(2(x-3))+x/(2x+2)=(2x)/((x+1)(x-3))(x ne -1,3)`
`<=>x(1/(2(x-3))+1/(2(x+1))-2/((x+1)(x-3)))=0`
`+)x=0`
`+)1/(2(x-3))+1/(2(x+1))-2/((x+1)(x-3))=0`
`<=>x+1+x-3-4=0` `<=>2x-6=0` `<=>x=3(l)`
Vậy `x=0`
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