Sơ đồ mạch điện: $(R_{1}//R_{2}//R_{3})nt(R_{4}//R_{5})$
$R=\dfrac{1}{\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}}+\dfrac{R_{4}.R_{5}}{R_{4}+R_{5}}$
$=\dfrac{153}{62}$ Ω
$⇒I=\dfrac{U}{R}=\dfrac{6}{\dfrac{153}{62}}=\dfrac{124}{51}A$
Ta có: $R_{123}=\dfrac{1}{\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}}=1,5$ Ω
$R_{45}=\dfrac{R_{4}.R_{5}}{R_{4}+R_{5}}=\dfrac{30}{31}$ Ω
$⇒\dfrac{U_{1}}{U_{4}}=\dfrac{R_{123}}{R_{45}}$
$⇔\dfrac{6-U_{4}}{U_{4}}=\dfrac{1,5}{\dfrac{30}{31}}$
$⇔U_{4}=\dfrac{40}{17}V$
$⇒U_{1}=\dfrac{62}{17}V$
Có $U$ với $R$ rồi bạn tự tính $I$ nha