Tìm các số hữu tỉ \(x, y, z\) biết rằng: \(x(x + y + z) = -5; y(x + y + z) = 9;\)\(\, z(x + y + z) = 5.\)

\(x(x + y + z) = -5\)      (1)

\(y(x + y + z) = 9\)          (2)

\(\, z(x + y + z) = 5\)          (3)

Cộng theo từng vế các đẳng thức (1), (2), (3), ta được:

\(x(x + y + z) +y(x + y + z) \)\(\,+ z(x + y + z) = -5+9+5\)

\((x+y+z).(x+y+z)=9\)

\({\left( {x + y + z} \right)^2} = 9\)

\(\Rightarrow x + y + z =  \pm 3\) 

+) Nếu \(x + y + z = 3\) thì

\(\begin{array}{l}
x = \dfrac{{x\left( {x + y + z} \right)}}{{x + y + z}} = \dfrac{{ - 5}}{3}\\
y = \dfrac{{y\left( {x + y + z} \right)}}{{x + y + z}} = \dfrac{9}{3} = 3\\
z = \dfrac{{z\left( {x + y + z} \right)}}{{x + y + z}} = \dfrac{5}{3}
\end{array}\)

Vậy \(\displaystyle x = {{ - 5} \over 3},y = 3,z = {5 \over 3}\)

+) Nếu \(x + y + z = -3\) thì 

\(\begin{array}{l}
x = \dfrac{{x\left( {x + y + z} \right)}}{{x + y + z}} = \dfrac{{ - 5}}{{ - 3}} = \dfrac{5}{3}\\
y = \dfrac{{y\left( {x + y + z} \right)}}{{x + y + z}} = \dfrac{9}{{ - 3}} = - 3\\
z = \dfrac{{z\left( {x + y + z} \right)}}{{x + y + z}} = \dfrac{5}{{ - 3}} = \dfrac{{ - 5}}{3}
\end{array}\)

Vậy \(\displaystyle x = {5 \over 3},y =  - 3,z = {{ - 5} \over 3}\)