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[latex] \left\{\begin{array}{c}\frac{1}{x+y}+\frac{1}{y+z}=1,5\\\frac{1}{y+z}+\frac{1}{z+x}=0,7\\\frac{1}{z+x}+\frac{1}{x+y}=1,2\end{array}\right \; \; ;\; \; a=\frac{1}{x+y}\; ,\; \; b=\frac{1}{y+z}\; ,\; \;c=\frac{1}{z+x}\\\\\\ \left\{\begin{array}{c}a+b=1,5\\b+c=0,7\\c+a=1,2\end{array}\right \; \oplus \; \; \; \; 2a+2b+2c=1,5+0,7+1,2\\\\\\2(a+b+c)=3,4\; \; \to \; \; \; a+b+c=1,7\\\\c=(a+b+c)-(a+b)=1,7-1,5=0,2\\\\a=(a+b+c)-(b+c)=1,7-0,7=1\\\\b=(a+b+c)-(c+a)=1,7-1,2=0,5[/latex] [latex] \left\{\begin{array}{c}\frac{1}{x+y}=1\\\frac{1}{y+z}=0,5\\\frac{1}{z+x}=0,2\end{array}\right\; \; \left\{\begin{array}{c}x+y=1\\y+z=\frac{1}{0,5}\\z+x=\frac{1}{0,2}\end{array}\right\; \; \left\{\begin{array}{ccc}x+y=1\\y+z=2\\z+x=5\end{array}\right\; \; \oplus \\\\\\2(x+y+z)=8\; \; \to \; \; \; x+y+z=4\\\\x=(x+y+z)-(y+z)=4-2=2\\\\y=(x+y+z)-(z+x)=4-5=-1\\\\z=(x+y+z)-(x+y)=4-1=3\\\\Otvet:\; \; (2;-1;3)\; .[/latex]