Ta có \(1-\dfrac{1}{2+\dfrac{1}{3}}=\dfrac{4}{7}=\dfrac{1}{\dfrac{7}{4}}=\dfrac{1}{1+\dfrac{3}{4}}=\dfrac{1}{1+\dfrac{1}{\dfrac{4}{3}}}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3}}}\)
Thay vào, ta có \(\dfrac{1}{x+\dfrac{1}{y+\dfrac{1}{z}}}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3}}}\Leftrightarrow x=1,y=1,z=3\left(tm\right)\)
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