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Ta có :
\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
Vậy \(a=1;b=2;c=3;d=4\)
Ta có: \(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
\(\Rightarrow\)a = 1 ; b = 2 ; c = 3 ; d = 4
Vậy:
a = 1 ; b = 2 ; c = 3 ; d = 4
\(\frac{30}{43}=\frac{1}{a+\frac{1}{b+\frac{1}{\frac{cd+1}{d}}}}=\frac{1}{a+\frac{1}{b+\frac{d}{cd+1}}}=\frac{1}{a+\frac{1}{\frac{bcd+b+d}{cd+1}}}\)
\(\frac{30}{43}=\frac{1}{a+\frac{cd+1}{bcd+b+d}}=\frac{1}{\frac{abcd+ab+ad+cd+1}{bcd+b+d}}\)
\(\frac{30}{43}=\frac{bcd+b+d}{abcd+ab+ad+cd+1}\)
\(a=1;b=2;c=3;d=4\)
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
\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
=> a = 1; b = 2; c = 3; d = 4.
\(\frac{30}{43}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{\frac{43}{30}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)
\(\Rightarrow\frac{1}{1+\frac{13}{30}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)
\(\Rightarrow a=1,b=2,c=3,d=4\)
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