Question

determianti numerele naturale a,b,c stiind ca 20 (a+b)=15(b+c)=12(c+a) si a la puterea 2 cu b putera 3 , c =559872

Answer 1

 

$\displaystyle\bf\\\text{Consider ca ecuatia a 3-a este: }~~a^2\times b^3\times c=559872\\\\Avem~ecuatiile:\\\\20(a+b)=15(b+c)=12(c+a)\\a^2\times b^3\times c=559872\\\text{Sunt 3 ecuatii cu 3 necunoscute din care 2 ecuatii sunt liniare.}\\\text{La inceput ne vom ocupa doar de ecuatiile liniare.}$

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$\displaystyle\bf\\Rezolvare:\\\\20(a+b)=15(b+c)=12(c+a)\\\\\begin{cases}\bf20(a+b)=15(b+c)\\\bf15(b+c)=12(c+a)\end{cases}\\\\\text{Avem 2 ecuatii cu 3 necunoscute.}\\\text{Il trecem pe c in dreapta si le calculam pe a si pe b in functie de c.}\\\\\begin{cases}\bf20a+20b=15b+15c\\\bf15b+15c)=12c+12a\end{cases}\\\\\begin{cases}\bf20a+20b-15b=15c\\\bf-12a +15b = 12c - 15c\end{cases}\\\\\begin{cases}\bf20a+5b=15c~~~\Big|:5\\\bf-12a +15b = -3c~~~\Big|:3\end{cases}$

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$\displaystyle\bf\\ \begin{cases}\bf4a+b=3c\\\bf-4a +5b = -c\end{cases}\\ ---------~~~Adunam~ecuatiile.\\\\~~~~~/~~~6b=2c\\\\b=\frac{2c}{6}\\\\\boxed{\bf~b=\frac{c}{3}}\\\\4a+b=3c\\4a=3c-b\\\\4a=3c-\frac{c}{3}\\\\4a=\frac{9c-c}{3}\\\\4a=\frac{8c}{3}~~\Big|:4\\\\\boxed{\bf~a=\frac{2c}{3}}\\\\\boxed{\bf~c=c}$

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$\displaystyle\bf\\Rezumat:\\\\\begin{cases}\bf~b=\dfrac{c}{3}\\\\\bf~a=\dfrac{2c}{3}\\\\\bf~c=c\end{cases}\\\\\\\text{Mergem la ecuatia a 3-a}\\\text{Descompunem numarul 559872 in factori primi.}\\\\559872=2^8\times3^7\\\\a^2\times b^3\times c = 559872\\\\\left(\frac{2c}{3}\right)^2\times\left(\frac{c}{3}\right)^3\times c=2^8\times3^7\\\\\frac{2^2c^2}{3^2}\times\frac{c^3}{3^3}\times c=2^8\times3^7\\\\\\\frac{2^2c^2\times c^3\times c}{3^2\times3^3}=2^8\times3^7$

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$\displaystyle\bf\\\frac{2^2c^6}{3^5}=2^8\times3^7\\\\\\c^6=\frac{2^8\times3^7\times3^5}{2^2}\\\\\\c^6=2^{8-2}\times3^{7+5}\\\\c^6=2^6\times3^{12}\\\\c^6=2^6\times3^{2\times6}\\\\c^6=2^6\times\Big(3^2\Big)^6\\\\c^6=\Big(2\times3^2\Big)^6\\\\\implies~c=2\times3^2\\\boxed{\bf~c=18}\\\\a=\frac{2\times18}{3}=\frac{36}{3}\\\\\boxed{\bf~a=12}\\\\b=\frac{c}{3}=\frac{18}{3}\\\\\boxed{\bf~b=6}$