Với `x \ne 0,x \ne 3`. Ta có:
`x / [ x - 3 ] : ( x / [ x^2 - 3x ] - 2 / x )`
`= x / [ x - 3 ] : ( x / [ x ( x - 3 ) ] - [ 2 ( x - 3 ) ] / [ x ( x - 3 ) ] )`
`= x / [ x - 3 ] : [ x - 2x + 6 ] / [ x ( x - 3 ) ]`
`= x / [ x - 3 ] . [ x ( x - 3 ) ] / [ -x + 6 ]`
`= [ x^2 ] / [ -x + 6 ]`