Question

giải thích thật rõ cho mk vs nhé!

Answer 1

Ta có:

\(^{2b^2}=c^2+28\)

<=>\(2b^2+c^2=28\)

Theo đề bài,ta có:

\(\frac{a}{6}=\frac{b}{8}=\frac{c}{11}\)=>\(\frac{a^2}{6^2}=\frac{b^2}{8^2}=\frac{c^2}{11^2}\)=\(\frac{a^2}{36}=\frac{b^2}{64}=\frac{c^2}{121}\)=\(\frac{2b^2}{128}\)=\(\frac{2b^2+c^2}{128-121}\)=\(\frac{28}{7}\)=4

Ta có:

\(\left\{\begin{matrix}\frac{a^2}{36}=4\\\frac{b^2}{64}=4\\\frac{c^2}{121}=4\end{matrix}\right.=>\left\{\begin{matrix}a^2=144\\b^2=256\\c^2=484\end{matrix}\right.=>\left\{\begin{matrix}a=12\\b=16\\b=22\end{matrix}\right.\)

Vậy:

\(\left\{\begin{matrix}a=12\\b=16\\c=22\end{matrix}\right.\)