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`d, 5x(2x-1/2)+(4x-1)=0`

`5x(1/2 . 4x -1/2 . 1) +(4x-1)=0`

`5x . 1/2(4x -1)+(4x-1)=0`

`5/2x(4x-1)+1(4x-1)=0`

`(5/2x+1)(4x-1)=0`

`5/2x+1=0` hoặc `4x-1=0`

`5/2x=-1` hoặc `4x=1`

`x=-2/5` hoặc `x=1/4`

Vậy `x ∈ {-2/5;1/4}`