Question

Indicați amplificarile și/sau simplificările necesare pentru a obține fracția 15/35 plecând de la fracția. Va rog eu mult ​

Answer 1

 

Amplificam si/sau smplificam fractiile urmatoare pentru a obtine 15/35.

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$\displaystyle\bf\\d)~~\frac{1515}{3535}=\frac{15\times100+15}{35\times100+35}=\frac{15(100+1)}{35(100+1)}=\frac{15\times101}{35\times101}=\frac{15}{35}\\\\Am~simplificat~cu~101\\\\\\e)~~\frac{9x+9}{21x+21}=\frac{3(3x+3)}{7(3x+3)}=\frac{^{5)}3}{~~7}=\frac{15}{35}\\\\Am~simplificat~cu~(3x+3)~si~am~amplificat~cu~5$

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$\displaystyle\bf\\f)~~\frac{3a+3}{7a+7}=\frac{3(a+1)}{7(a+1)}=\frac{^{5)}3}{~~7}=\frac{15}{35}\\\\Am~simplificat~cu~(a+1)~si~am~amplificat~cu~5\\\\\\g)~~\frac{2^{n+2}-2^n}{2^{n+2}+2^{n+1}+2^n}=\frac{2^n\times2^2-2^n}{2^n\times2^2+2^n\times2^1+2^n}=\\\\=\frac{2^n(2^2-1)}{2^n(2^2+2^1+1)}=\frac{2^n\times3}{2^n\times7}=\frac{^{5)}3}{~~7}=\frac{15}{35}\\\\Am~simplificat~cu~2^n~si~am~amplificat~cu~5$

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$\displaystyle\bf\\h)~~\frac{5^{n+1}-2\cdot5^n}{5^{n+1}+2\cdot5^n}=\frac{5^n\cdot5^1-2\cdot5^n}{5^n\cdot5+2\cdot5^n}=\frac{5^n(5-2)}{5^n(5+2)}=\frac{5^n\cdot3}{5^n\cdot7} =\frac{^{5)}3}{~~7}=\frac{15}{35}\\\\Am~simplificat~cu~5^n~si~am~amplificat~cu~5.$