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\(1+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}=\frac{43}{30}\)
\(\frac{1}{x+\frac{1}{y+\frac{1}{z}}}=\frac{43}{30}-1\)
\(\frac{1}{x+\frac{1}{y+\frac{1}{z}}}=\frac{13}{30}\)
Ta có:\(\frac{13}{30}=\frac{1}{\frac{30}{13}}=\frac{1}{2+\frac{4}{13}}=\frac{1}{2+\frac{1}{\frac{13}{4}}}=\frac{1}{2+\frac{1}{3+\frac{1}{4}}}\)
Vậy x=2; y=3; z=4
ta có:
\(\frac{43}{30}=1+\frac{13}{30}=1+\frac{1}{\frac{30}{13}}=1+\frac{1}{2+\frac{4}{13}}\)\(=1+\frac{1}{2+\frac{1}{\frac{13}{4}}}=1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}=1+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}\)
=>x=2;y=3;z=4
Vậy (x;y;z)=(2;3;4)
\(\frac{43}{30}=1+\frac{13}{30}=1+\frac{1}{\frac{30}{13}}=1+\frac{1}{2+\frac{4}{13}}=1+\frac{1}{2+\frac{1}{\frac{13}{4}}}=1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}\)
Do đó ta có: \(1+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}=1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}\)
=>x=2;y=3;z=4
Ta có:
\(\frac{43}{30}=1+\frac{13}{30}=1+\frac{1}{\frac{30}{13}}=1+\frac{1}{2+\frac{4}{13}}=1+\frac{1}{2+\frac{1}{\frac{13}{4}}}=1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}\)
Thay vào x,y,z vậy x=2;y=3;z=4
Chúc bạn học tốt!
Đề bé tý: số hữu Tỷ: \(\frac{43}{30}=1+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}\)
\(\frac{43}{30}=1+\frac{13}{30}=1+\frac{1}{2+\frac{4}{13}}=1+\frac{1}{2+\frac{1}{\frac{13}{4}}}=1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}\)
Vậy (x,y,z)=(2,3,4)
\(\frac{43}{30}=1+\frac{13}{30}\)
hay \(\frac{1}{x+\frac{1}{y+\frac{1}{z}}}\)= \(\frac{13}{30}\)
=> \(x+\frac{1}{y+\frac{1}{z}}\) = \(\frac{30}{13}=2+\frac{4}{13}\) => x = 2.
=> \(\frac{1}{y+\frac{1}{z}}\) = \(\frac{4}{13}\) => \(y+\frac{1}{z}\) = \(\frac{13}{4}\) = \(3+\frac{1}{4}\) => y = 3, z = 4.
Vậy x = 2, y = 3, z = 4.
mk nha =))
Ta có:
\(\frac{43}{30}=1+\frac{13}{30}\)
\(\Rightarrow\frac{1}{x+\frac{1}{y\frac{1}{x}}}=\frac{13}{30}\)
hay \(x+\frac{1}{y+\frac{1}{x}}=\frac{30}{13}\)
mà \(\frac{30}{13}=2+\frac{4}{13}\Rightarrow x=2\)
\(\Rightarrow\frac{1}{y+\frac{1}{z}}=\frac{4}{13}\)
hay \(y+\frac{1}{z}=\frac{13}{4}=3+\frac{1}{4}\Rightarrow y=3\)
\(\Rightarrow z=4\)
Vậy \(x=2;y=3;z=4\)