sistem

[ghenghea1]

Să se rezolve sistemul: $\left\{\begin{array}{c} \dfrac{2x-y+z-1}{2x-y+z} - \dfrac{1}{x+y+z}+\dfrac{2}{x+y-z}=\dfrac{3}{4}\\ \dfrac{1}{2x-y+z}-\dfrac{x+y+z+1}{x+y+z}+\dfrac{1}{x+y-z}=\dfrac{1}{4}\\ \dfrac{2}{2x-y+z}+\dfrac{4}{x+y+z}+\dfrac{x-y+z+2}{x-y+z}=5\end{array}\right$.

[mihai miculita]

Sistemul dat poate fi adus la forma:
$\left\{\begin{array}{c} -\dfrac{1}{2x-y+z}-\dfrac{1}{x+y+z}+\dfrac{2}{x+y-z}=-\dfrac{1}{4}\\ \dfrac{1}{2x-y+z}-\dfrac{1}{x+y+z}+\dfrac{1}{x+y-z}=\dfrac{5}{4};\,\,\,(1)\\ \dfrac{2}{2x-y+z}-\dfrac{4}{x+y+z}+\dfrac{2}{x+y-z}=4\end{array}\right$
si facand acum substitutia: $\boxed{a=\dfrac{1}{2x-y+z};\,b=\dfrac{1}{x+y+z},c=\dfrac{1}{x+y-z}};\,\,\,(2),$ sistemul (1) devine:
$\left\{\begin{array}{c} -a-b+2c=-\dfrac{1}{4}\\ a-b+c=\dfrac{5}{4}\\ 2a+4b+2c=4\end{array}\right$$\Rightarrow\left\{\begin{array}{c} -2b+3c=1\\ 2b+6c=\dfrac{7}{2}\end{array}\right\Rightarray 9c=\dfrac{9}{2}\Rightarrow \boxed{c=\dfrac{1}{2}}.$
Asa ca, din:
$\left\{\begin{array}{c} c=\dfrac{1}{2}\\ 2b+6c=\dfrac{7}{2}\end{array}\right \Rightarrow 2b+3=\dfrac{7}{2}\Rightarrow \boxed{b=\dfrac{1}{4}}.$
In fine, din:
$\left \begin{array}{c} a-b+c=\dfrac{5}{4}\\ b=\dfrac{1}{4}\\ c=\dfrac{1}{2}\end{array}\right\}\Rightarrow a=b-c+\dfrac{5}{4}=\dfrac{1}{4}-\dfrac{1}{2}+\dfrac{5}{4}=1\Rightarrow \boxed{a=1}.$
Inlocuid acum valorile $a=1,b=\dfrac{1}{4}$ si $c=\dfrac{1}{2}$ in relatiile (2) ajungem la sistemul urmator:
$\left\{\begin{array}{c} 2x-y+z=2\\ x+y+z=4\\ x-y+z=1\end{array}\right;$ care are solutia:$\boxed{ \left(x;y;z\right)=\left(1;\dfrac{3}{2};\dfrac{3}{2}\right)}.$