Question

Answer 1

\(\left(x+\dfrac{1}{12}\right)^2=\dfrac{16}{9}\\ =>\left(x+\dfrac{1}{12}\right)^2=\left(\dfrac{4}{3}\right)^2\\ TH1:x+\dfrac{1}{12}=\dfrac{4}{3}\\ =>x=\dfrac{4}{3}-\dfrac{1}{12}\\ =>x=\dfrac{15}{12}=\dfrac{5}{4}\\ TH2:x+\dfrac{1}{12}=\dfrac{-4}{3}\\ =>x=\dfrac{-4}{3}-\dfrac{1}{12}\\ =>x=\dfrac{-17}{12}\)

Answer 2

:)))

Answer 3

`(x + 1/12)^2 = 16/9`

`=> (x + 1/12)^2 = (+-4/3)^2`

`=> x+ 1/12 = +-4/3`

Trường hợp `1:`

`x + 1/12 = 4/3`

`=> x = 5/4`

Trường hợp `2:`

`x + 1/12 = - 4/3`

`x  = -17/12`

Vậy `x in {5/4 ; -17/12}`