Răspuns: [tex]\bf \dfrac{2}{15} \implies fractie\:subunitara[/tex]
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[tex]\bf \dfrac{3^{x+1}+3^{x+2}}{3^{x+2}+3^{x+4}} =\dfrac{3^{x+1}\cdot(3^{x+1-(x+1)}+3^{x+2-(x+1)})}{3^{x+2}\cdot(3^{x+2-(x+2)}+3^{x+4-(x+2)})}=\dfrac{3^{x+1}\cdot(3^{0}+3^{1})}{3^{x+2}\cdot(3^{0}+3^{2})}=[/tex]
[tex]\bf \dfrac{3^{x+1}\cdot(1+3)}{3^{x+2}\cdot(1+9)}=\dfrac{3^{x}\cdot3^{1}\cdot 4}{3^{x}\cdot3^{2}\cdot 10}=\dfrac{\not3^{x}\cdot3^{1}\cdot \not4}{\not3^{x}\cdot3^{2}\cdot \not10}= \dfrac{3^{1}\cdot 2}{3^{2}\cdot 5}= \dfrac{\not3^{1}\cdot 2}{\not3^{2}\cdot 5}= \dfrac{2}{3\cdot 5}= \boxed{\bf \dfrac{2}{15}}[/tex]
[tex]\bf \dfrac{2}{15} \implies fractie\:subunitara[/tex]
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