Question

Cho x:y:z \(\in\)Q:thỏa mãn:

x+y=\(\frac{7}{12}\)

y+z=\(\frac{-19}{24}\)

z+x=\(\frac{1}{8}\)

Tìm x ;y ;z ?

Answer 1

\(x+y=\frac{7}{12}\Rightarrow x=\frac{7}{12}-y\)(1)\(y+z=\frac{-19}{24}\Rightarrow z=\frac{-19}{24}-y\)(2)\(z+x=\frac{1}{8}\)(3)

Từ (1),(2) và (3) \(\Rightarrow\left(\frac{-19}{24}-y\right)+\left(\frac{7}{12}-y\right)=\frac{1}{8}\)

\(\Rightarrow\frac{-19}{24}-y+\frac{7}{12}-y=\frac{1}{8}\)

\(\Rightarrow\frac{-5}{24}-2y=\frac{1}{8}\)

\(\Rightarrow-2y=\frac{1}{8}+\frac{5}{24}\)

\(\Rightarrow-2y=\frac{1}{3}\)

\(\Rightarrow y=\frac{1}{3}:\left(-2\right)\)

\(\Rightarrow y=-\frac{1}{6}\)

Vì \(x=\frac{7}{12}-y\)mà \(y=-\frac{1}{6}\);\(\Rightarrow x=\frac{7}{12}-\frac{-1}{6}=\frac{7}{12}+\frac{1}{6}=\frac{7+2}{12}=\frac{9}{12}=\frac{3}{4}\)

Vì \(z=\frac{-19}{24}-y\)mà \(y=-\frac{1}{6}\);\(\Rightarrow z=\frac{-19}{24}-\frac{-1}{6}=\frac{-19}{24}+\frac{1}{6}=\frac{-19+4}{24}=\frac{-15}{24}=\frac{-5}{8}\)

Vậy

\(x=\frac{3}{4}\)\(y=-\frac{1}{6}\)\(z=-\frac{5}{8}\)

Answer 2

\(x=\frac{3}{4}\)

\(y=\frac{-1}{6}\)

\(z=\frac{-5}{8}\)