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\( \frac{1}{x} + \frac{1}{y+z}=\frac{1}{2} \)

\( \frac{1}{y} + \frac{1}{z+x}=\frac{1}{3} \)

\( \frac{1}{z} + \frac{1}{x+y}=\frac{1}{4} \)



Mie mi-a dat o singura solutie: (x,y,z)={(2,3;3,8(3);11,5)}
Mai sunt alte solutii? (+rezolvarea problemei).
Multumesc.

E bine.

\( \left\{\begin{array}{c} \frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}\\
\frac{1}{y}+\frac{1}{x+z}=\frac{1}{3}\\
\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4}\end{array} \Rightarrow

\left\{\begin{array}{c} \frac{x+y+z}{x(y+z)}=\frac{1}{2}\\
\frac{x+y+z}{y(x+z)}=\frac{1}{3}\\
\frac{x+y+z}{z(x+y)}=\frac{1}{4}\end{array} \Rightarrow
\frac{x(y+z)}{2}=\frac{y(x+z)}{3}=\frac{z(x+y)}{4}=x+y+z.\ \ (1)
\)
\( \frac{x(y+z)}{2}=\frac{y(x+z)}{3}=\frac{z(x+y)}{4}\Rightarrow
\left\{\begin{array}{c} xy+3xz-2yz=0\\
2xy+xy-yz=0 \end{array} \Rightarrow

\left\{\begin{array}{c} \frac{1}{z}+\frac{3}{y}-\frac{2}{x}=0\\
\frac{2}{z}+\frac{1}{y}-\frac{1}{x}=0 \end{array}\Rightarrow
\frac{x}{3}=\frac{y}{5}=\frac{z}{15}=k\Rightarrow \left\{ \begin{array}{c} x=3k\\
y=5k\\
z=15k\end{array}\ \ (2) \)
\(
\mbox{In fine, din (1) si (2) obtinem: } k=\frac{23}{30}\Rightarrow \left\{\begin{array} x=\frac{23}{10}=2,3\\
y=\frac{23}{6}=3,8(3)\\
z=\frac{23}{2}=11,5.\end{array} \)