`#tnvt`
`a)\frac{5}{x-3}=\frac{x+4}{12}(x\ne3)`
`=>5.12=(x-3).(x+4)`
`=>x^2+x-12=60`
`=>x^2+x-72=0`
`=>x^2+9x-8x-72=0`
`=>(x+9)(x-8)=0`
`=>[(x=-9(tm)),(x=8(tm)):}`
Vậy `x\in{-9;8}`
`b)\frac{x+3}{12}=\frac{3}{x+3}(x\ne-3)`
`=>(x+3)(x+3)=3.12`
`=>(x+3)^2=36`
`=>(x+3)^2=(+-6)^2`
`=>[(x+3=6),(x+3=-6):}`
`=>[(x=3(tm)),(x=-9(tm)):}`
Vậy `x\in{3;-9}`
`c)2x^2+(-6)^3 :27=0`
`=>2x^2+(-216):27=0`
`=>2x^2+(-8)=0`
`=>2x^2=8`
`=>x^2=4`
`=>x^2=(+-2)^2`
`=>x^2=+-2`
Vậy `x\in{+-2}`
