Đáp án:
`a=82436;b=4;c=5;d=1;e=17/2`
Giải thích các bước giải:
$\text{Ta có}:\dfrac{20032004}{243}\\=82436+\dfrac{56}{243}\\=82436+\dfrac{1}{\dfrac{243}{56}}\\=82436+\dfrac{1}{4+\dfrac{19}{56}}\\=82436+\dfrac{1}{4+\dfrac{2}{\dfrac{112}{19}}}\\=82436+\dfrac{1}{4+\dfrac{2}{5+\dfrac{17}{19}}}\\=82436+\dfrac{1}{4+\dfrac{2}{5+\dfrac{1}{\dfrac{19}{17}}}}\\=82436+\dfrac{1}{4+\dfrac{2}{5+\dfrac{1}{1+\dfrac{2}{17}}}}\\=82436+\dfrac{1}{4+\dfrac{2}{5+\dfrac{1}{1+\dfrac{1}{\dfrac{17}{2}}}}}$
Vậy `a=82436;b=4;c=5;d=1;e=17/2`
