`c ) ( x + 1/2 )^2 = 1/16`
Trường hợp `1 : ( x + 1/2 )^2 = ( 1/4 )^2`
`⇔ x + 1/2 = 1/4`
`⇔ x = 1/4 - 1/2`
`⇔ x = ( - 1 )/4`
Trường hợp `2 : ( x + 1/2 )^2 = ( ( - 1 )/4 )^2`
`⇔ x + 1/2 = ( - 1 )/4`
`⇔ x = ( - 1 )/4 - 1/2`
`⇔ x = ( - 3 )/4`
Vậy `, x ∈ { ( - 1 )/4 ; ( - 3 )/4 } .`
`d, ( 3x + 1 )^3 = - 27`
`⇔ ( 3x + 1 )^3 = ( - 3 )^3`
`⇔ 3x + 1 = - 3`
`⇔ 3x = - 3 - 1`
`⇔ 3x = - 4`
`⇔ x = - 4 : 3`
`⇔ x = ( - 4 )/3`
Vậy `, x = ( - 4 )/3 .`
