Question

Answer 1

  

Answer 2

\(\left(x-\dfrac{5}{6}\right)^2=\left(\dfrac{1}{6}\right)^2\)

\(x-\dfrac{5}{6}=\dfrac{1}{6}\)

\(x=\dfrac{1}{6}+\dfrac{5}{6}\)

\(x=1\)

Vậy x = 1

Answer 3

\(\left(x-\dfrac{5}{6}\right)^2=\dfrac{1}{36}\\ =>\left(x-\dfrac{5}{6}\right)^2=\left(\dfrac{1}{6}\right)^2\\TH1:x-\dfrac{5}{6}=\dfrac{1}{6}\\ =>x=\dfrac{1}{6}+\dfrac{5}{6}\\ =>x=\dfrac{6}{6}=1\\ TH2:x-\dfrac{5}{6}=-\dfrac{1}{6}\\ =>x=\dfrac{-1}{6}+\dfrac{5}{6}\\ =>x=\dfrac{4}{6}=\dfrac{2}{3}\)

Answer 4

`( x - 5/6)^2 = 1/36`

`=> (x - 5/6)^2 = (+-1/6)^2`

`=> x - 5/6 = +-1/6`

Trường hợp `1:`

`x - 5/6 = 1/6`

`=> x = 1`

Trường hợp `2:`

`x - 5/6 = - 1/6`

`=> x = 2/3`

Vậy `x in {1;2/3}`