`a)`
`(5x+1)^2 = 36/49`
`=> (5x+1)^2 = (6/7)^2`
`=> 5x + 1= 6/7` hoặc `5x+1=-6/7`
`+) 5x + 1 =6/7`
`=> 5x = -1/7`
`=> x = -1/35`
`+) 5x +1 = -6/7`
`=> 5x = -13/7`
`=> x = -13/35`
Vậy `x \in {-1/35 ; -13/35}`
`b)`
`(x-2/9)^3 = (2/3)^6`
` =>(x-2/9)^3 = [(2/3)^2]^3`
`=> (x-2/9)^3 = (4/9)^3`
`=> x - 2/9 = 4/9`
`=> x = 4/9 + 2/9`
`=> x = 2/3`
Vậy `x=2/3`