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\(\frac{2003}{273}=7+\frac{92}{273}=7+\frac{1}{\frac{273}{92}}=7+\frac{1}{2+\frac{89}{92}}=7+\frac{1}{2+\frac{1}{\frac{92}{89}}}\)
\(=7+\frac{1}{2+\frac{1}{1+\frac{3}{89}}}=7+\frac{1}{2+\frac{1}{1+\frac{1}{\frac{89}{3}}}}\)
\(=7+\frac{1}{2+\frac{1}{1+\frac{1}{29+\frac{2}{3}}}}=7+\frac{1}{2+\frac{1}{1+\frac{1}{29+\frac{1}{\frac{3}{2}}}}}\)
\(=7+\frac{1}{2+\frac{1}{1+\frac{1}{29+\frac{1}{1+\frac{1}{2}}}}}\)
Do đó \(a=1;b=29;c=1;d=2\)
\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
Vay a=1; b=2; c=3; d=4
Ta có : \(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
Vậy a = 1,b = 2,c = 3,d = 4
\(\frac{30}{43}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{\frac{43}{30}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)
\(\Rightarrow\frac{1}{1+\frac{13}{30}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)
\(\Rightarrow a=1,b=2,c=3,d=4\)