$R_1=2R_2=3R_3⇒R_2=\dfrac{1}{2}R_1,R_3=\dfrac{1}{3}R_1$

Vì $R_1$ // $R_2$ // $R_3$

Nên $R_{tđ}=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}$

$⇔R_{tđ}=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{\frac{1}{2}R_1}+\dfrac{1}{\frac{1}{3}R_1}}$

$⇔R_{tđ}=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{2}{R_1}+\dfrac{3}{R_1}}$

$⇔\dfrac{U}{I}=\dfrac{1}{\dfrac{6}{R_1}}$

$⇔\dfrac{U}{I}=\dfrac{R_1}{6}$

$⇔R_1=6\dfrac{U}{I}$

$T/s:R_1=6.\dfrac{36}{2}=108\space(\Omega)$

      $⇒R_2=\dfrac{1}{2}R_1=\dfrac{1}{2}.108=54\space(\Omega)$

           $R_3=\dfrac{1}{3}R_1=\dfrac{1}{3}.108=36\space(\Omega)$

Vậy $R_1=108\space\Omega, R_2=54\space\Omega,R_3=36\space\Omega$

Đáp án `+` Giải thích các bước giải:

`R_1 = 2R_2 ⇔ R_2 = R_1/2`

`R_1 = 3R_3 ⇔ R_3 = R_1/3`

Điện trở tương đương : 

`R = U/I = 36/2 = 18(\Omega)`

Ta có : `R = (R_1 R_2 R_3)/( R_1 R_3 + R_2 R_3 + R_2 R_1)`

$⇔ \dfrac{R_1 \dfrac{R_1}{2}  \dfrac{R_1}{3}}{R_1 \dfrac{R_1}{3} + \dfrac{R_1}{2}  \dfrac{R_1}{3} + \dfrac{R_1}{2} R_1} = 18$

$⇔ \dfrac{\dfrac{R_1^3}{6}}{\dfrac{R_1^2}{3} + \dfrac{R_1^2}{6} + \dfrac{R_1^2}{2}}=18$

$⇔ \dfrac{\dfrac{R_1^3}{6}}{R_1^2}=18$

$⇔R_1 =108(\Omega)$

`R_2 = R_1/2 = 108/2 = 54(\Omega)`

`R_3 = R_1/3 =108/3 = 36(\Omega)`.