Răspuns :
Folosim proprietatea fundamentala a proportiei
[tex]\dfrac{3+x}{8}=\dfrac{1}{3-x}\Rightarrow (3-x)(3+x)=8\Rightarrow9-x^2=8\Rightarrow x^2=1[/tex]
[tex]\Rightarrow x_{1;2}=\pm1[/tex]
[tex]\dfrac{3+x}{8}=\dfrac{1}{3-x}\Rightarrow (3-x)(3+x)=8\Rightarrow9-x^2=8\Rightarrow x^2=1[/tex]
[tex]\Rightarrow x_{1;2}=\pm1[/tex]