`a) (2x-1)^3=(2x-1)^2`
`<=> (2x-1)^3-(2x-1)^2=0`
`<=> (2x-1)^2 . [(2x-1)-1]=0`
`<=> (2x-1)^2 . (2x-1-1)=0`
`<=> (2x-1)^2 . (2x-2)=0`
`<=> (2x-1)^2=0` hay `2x-2=0`
`<=> 2x-1=0` hay `2x=0+2`
`<=> 2x=0+1` hay `2x=2`
`<=> 2x=1` hay `x=2:2`
`<=> x=1/2` hay `x=1`
Vậy `x \in {1/2;1}`
`b) (1/2-4x)^4=(4x-1/2)^4`
`<=> [-1(4x-1/2)]^4=(4x-1/2)^4`
`<=> (-1)^4 . (4x-1/2)^4=(4x-1/2)^4`
`<=> 1 . (4x-1/2)^4=(4x-1/2)^4`
`<=> (4x-1/2)^4=(4x-1/2)^4`
`<=> (4x-1/2)^4-(4x-1/2)^4=0`
`<=> 0=0` ( luôn đúng )
Vậy biểu thức thỏa mãn với mọi `x`
