$a)$$R_t$$_đ$$=$$R_1$$+($$\dfrac{R_2.R_3}{R_2+R_3}$$)$$=18+($$\dfrac{30.20}{30+20}$$)$$=30($$\Omega$$)$

$b)$$I=$$\dfrac{U}{R_{tđ}}$$=$$\dfrac{60}{30}$$=2(A)$
$I=$$I_1$$=$$I_{23}$$=2A($$R_1$$nt$$R_{23}$$)$

$U_1$$=$$I_1$$.$$R_1$$=2.18=36(V)$

$\Rightarrow$ $U_{23}$$=$$U_2$$=$$U_3$$=$$U-$$U_1$$=60-36=24(V)($$R_2$//$R_3$$)$

$\begin{cases} I_2=\dfrac{U_2}{R_2}=\dfrac{24}{30}=0,8(A)\\I_3=\dfrac{U_3}{R_3}=\dfrac{24}{20}=1,2(A) \end{cases}$

@Nobitao@

a) Rtđ = R1+ [(R2 x R3) / (R2+R3)] =18+[(30 x 20) / (30+20)]= 30Ω

b) Uac = I x Rtđ  (với I là I toàn mạch)

   => I = Uac / Rtđ = 60/30 = 2A

   Mà I = I1 => I1 = 2A

   Vì R2 // R3 => U2 = U3 = Ubc

   Lại có: Ubc = Uac - Uab. Mà Uab = I1 x R1= 2 x 18 = 36V

   => Ubc = 60 - 36 = 24V

Vậy I2 = Ubc / R2 = 24 / 30 = 0,8A
       I3 = Ubc / R3 = 24 / 20 =1,2A