Question

[Đỗ Ngọc Linh]

Số hữu tỉ \(\frac{43}{30}\)có thể viết dưới dạng 1 + \(\frac{1}{x+\frac{1}{y+\frac{1}{z}}}\). Tìm x ;y; z

Answer 1

[Kẻ Huỷ Diệt]

\(\frac{43}{30}=1+\frac{13}{30}\)                               

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 =))

Answer 2

[Đặng Tiến]

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 \(x=2;y=3;z=4\)