$R_{0}=4$ ôm

$R_{2}=15$ ôm

$Đ:6V-3W⇒R_{Đ}=12$ ôm

$U_{V}=3V$

$U=24V$

Sđmđ: $[(R_{1}ntĐ)//(R_{2}ntR_{3}]ntR_{0}$

Gọi $R_{1}=x$

$R_{3}=y$

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Vì đèn sáng bình thường

$⇒U_{Đ}=6V$, $I_{1}=I_{Đ}=0,5A$

$⇒U_{1}=I_{1}.R_{1}=0,5x$

$⇒U_{23}=U_{1}+U_{Đ}=0,5x+6$

$⇒I_{23}=\dfrac{U_{23}}{R_{2}+R_{3}}=\dfrac{0,5x+6}{15+y}=\dfrac{x+12}{30+2y}$

Ta có: $I_{0}=I_{Đ}+I_{23}=0,5+\dfrac{x+12}{30+2y}=\dfrac{17+x+y}{30+2y}$

$U_{0}=U-U_{23}=24-0,5x-6=18-0,5x$

$⇒R_{0}=\dfrac{U_{0}}{I_{0}}=\dfrac{18-0,5x}{\dfrac{17+x+y}{30+2y}}$

$=\dfrac{(18-0,5x)(30+2y)}{17+x+y}=\dfrac{540+36y-15x-xy}{17+x+y}$

Mà $R_{0}=4$

$⇒\dfrac{540+36y-15x-xy}{17+x+y}=4$

$⇔540+36y-15x-xy=68+4x+4y$

$⇔19x-32y+xy-472=0$

$⇔x(19+y)=472+32y$

$⇔x=\dfrac{472+32y}{19+y}$

Ta có: $U_{1}+U_{V}+U_{3}=U_{23}$

$⇔0,5x+3+I_{23}.R_{3}=0,5x+6$

$⇔\dfrac{x+12}{30+2y}.y=3$

$⇔xy+12y=90+6y$

$⇔x=\dfrac{90-6y}{y}$

$⇒\dfrac{472+32y}{19+y}=\dfrac{90-6y}{y}$

$⇔(472+32y).y=(90-6y).(19+y)$

$⇔472y+32y²=1710-24y-6y²$

$⇔38y²+24y-1238=0$

$⇔y≈5,4$ ôm

$⇒x≈10,7$ ôm

Vậy $R_{1}=10,7$ ôm

$R_{3}=5,4$ ôm