Question

tìm a,b,c,d

\(\frac{2003}{273}=7+\frac{1}{2+\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}}\)

 

Answer 1

\(\frac{2003}{273}=7+\frac{92}{273}=7+\frac{1}{\frac{273}{92}}=7+\frac{1}{2+\frac{89}{92}}=7+\frac{1}{2+\frac{1}{\frac{92}{89}}}\)

\(=7+\frac{1}{2+\frac{1}{1+\frac{3}{89}}}=7+\frac{1}{2+\frac{1}{1+\frac{1}{\frac{89}{3}}}}\)

\(=7+\frac{1}{2+\frac{1}{1+\frac{1}{29+\frac{2}{3}}}}=7+\frac{1}{2+\frac{1}{1+\frac{1}{29+\frac{1}{\frac{3}{2}}}}}\)

\(=7+\frac{1}{2+\frac{1}{1+\frac{1}{29+\frac{1}{1+\frac{1}{2}}}}}\)

Do đó \(a=1;b=29;c=1;d=2\)

Answer 2

ai giúp mình với mình đang cần gấp!!@@