ta có \(\frac{52}{a-20}=\frac{39}{b-15}=\frac{13}{c-5}\)

=> \(\frac{a-20}{52}=\frac{b-15}{39}=\frac{c-5}{13}\)

=> \(\frac{a}{52}-\frac{20}{52}=\frac{b}{39}-\frac{15}{39}=\frac{c}{13}-\frac{5}{13}\)

=>\(\frac{a}{52}-\frac{5}{13}=\frac{b}{39}-\frac{5}{13}=\frac{c}{13}-\frac{5}{13}\)

=> \(\frac{a}{52}=\frac{b}{39}=\frac{c}{13}\)

=>\(\frac{a^2}{52^2}=\frac{b^2}{39^2}=\frac{c^2}{13^2}=\frac{bc}{39.13}=\frac{3}{3.13.13}=\frac{1}{13^2}\)

=>\(\hept{\begin{cases}a^2=\frac{1}{13^2}.52^2=4^2\\b^2=\frac{1}{13^2}.39^2=3^2\\c^2=\frac{1}{13^2}.13^2=162\end{cases}}\)

=>\(\hept{\begin{cases}a=4;-4\\b=3;-3\\c=1;-1\end{cases}}\)

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Ta có : \(\frac{a-20}{52}=\frac{b-15}{39}=\frac{c-5}{13}\)

=> \(\frac{a}{52}-\frac{20}{52}=\frac{b}{39}-\frac{15}{39}=\frac{c}{13}-\frac{5}{13}\)

=> \(\frac{a}{52}-\frac{5}{13}=\frac{b}{39}-\frac{5}{13}=\frac{c}{13}-\frac{5}{13}\)

=> \(\frac{a}{52}=\frac{b}{39}=\frac{c}{13}\)

Đặt \(\frac{a}{52}=\frac{b}{39}=\frac{c}{13}=k\)

=> \(\hept{\begin{cases}a=52k\\b=39k\\c=13k\end{cases}}\)

=> b.c = 39k.13k

=> 507k² = 3

=> k = \(\pm\frac{1}{13}\)

=> \(\hept{\begin{cases}a=52\cdot\frac{1}{13}=4\\b=39\cdot\frac{1}{13}=3\\c=13\cdot\frac{1}{13}=1\end{cases}}\)hoặc \(\hept{\begin{cases}a=52\cdot\left(-\frac{1}{13}\right)=-4\\b=39\cdot\left(-\frac{1}{13}\right)=-3\\c=13\cdot\left(-\frac{1}{13}\right)=-1\end{cases}}\)

Tại hạ xin test cách khác :))

\(\frac{52}{a-20}=\frac{39}{b-15}=\frac{13}{c-5}\)

\(=>\frac{a-20}{52}=\frac{b-15}{39}=\frac{c-5}{13}\)

\(=>\frac{a}{52}-\frac{5}{13}=\frac{b}{39}-\frac{5}{13}=\frac{c}{13}-\frac{5}{13}\)

\(=>\frac{a}{52}=\frac{b}{39}=\frac{c}{13}\)

\(=>\frac{a^2}{52^2}=\frac{b^2}{39^2}=\frac{c^2}{13^2}=\frac{bc}{39.13}=\frac{3}{13^2.3}=\frac{1}{13^2}\)

\(=>\hept{\begin{cases}\frac{a^2}{52^2}=\frac{1}{13^2}=>a^2=\frac{1}{13^2}.52^2=4^2\\\frac{b^2}{39^2}=\frac{1}{13^2}=>b^2=\frac{1}{13^2}.39^2=3^2\\\frac{c^2}{13^2}=\frac{1}{13^2}=>c^2=\frac{1}{13^2}.13^2=1^2\end{cases}}\)

\(=>\hept{\begin{cases}a=\pm4\\b=\pm3\\c=\pm1\end{cases}}\)

a=24 b=18c=5

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