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

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

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

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

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

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

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

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

a=24 b=18c=5

Ta có:

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

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

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

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

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

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

\(\Rightarrow\begin{cases}a^2=\frac{1}{13^2}.52^2=4^2\\b^2=\frac{1}{13^2}.39^2=3^2\\c^2=\frac{1}{13^2}.13^2=1^2\end{cases}\)\(\Rightarrow\begin{cases}a\in\left\{4;-4\right\}\\b\in\left\{3;-3\right\}\\c\in\left\{1;-1\right\}\end{cases}\)

Vậy giá trị (a;b;c) tương ứng thỏa mãn là: (4;3;1) ; (-4;-3;-1)

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