Với `x \ne +-2,x \ne 1/2,x \ne0`. Ta có:
`(3/[2x+4]+x/[2-x]+[2x^2+3]/[x^2-4]):[2x-1]/[4x-8]`
`=(3/[2(x+2)]-x/[x-2]+[2x^2+3]/[(x-2)(x+2)]).[4(x-2)]/[2x-1]`
`=[3(x-2)-2x(x+2)+2(2x^2+3)]/[x(x-2)(x+2)].[4(x-2)]/[2x-1]`
`=[3x-6-2x^2-4x+4x^2+6]/[x(x+2)]. 4/[2x-1]`
`=[2x^2-x]/[x(x+2)]. 4/[2x-1]`
`=[x(2x-1)]/[x(x+2)] . 4/[2x-1]`
`=4/[x+2]`