ta có 

\(-\frac{3^7\cdot2^8}{6^7}=-\frac{3^7\cdot2^7\cdot2}{6^7}=-\frac{6^7\cdot2}{6^7}=-2\)2

\(B=-\frac{3\cdot7^4+7^3}{7^5\cdot6-7^3\cdot2}=-\frac{3\cdot7^3\cdot7+7^3}{7^3\cdot7^2\cdot6-7^3\cdot2}=-\frac{7^3\cdot\left(3\cdot7+1\right)}{7^3\cdot\left(7^2\cdot6-2\right)}=-\frac{5}{73}\)

​\(\frac{x}{-3}=\frac{y}{7}\)và \(^{x^2-y^2=-360}\)

=> x=\(\frac{-3y}{7}\)

Thay vào: \(\left(\frac{-3}{7}.y\right)^2-y^2=-360\)

<=> \(\frac{-40}{49}.y^2=-360\)

<=>\(y^2=\frac{1}{441}\)

=> \(\orbr{\begin{cases}y=\frac{1}{21}\\y=\frac{-1}{21}\end{cases}}\)=>\(\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)

Đặt \(\frac{x}{-3}=\frac{y}{7}=k\)

\(\Rightarrow\hept{\begin{cases}x=-3k\\y=7k\end{cases}}\)

Thay vào ta có :

\(\left(-3k\right)^2-\left(7k\right)^2=-360\)

\(\left(-3\right)^2.k^2-7^2.k^2=-360\)

\(9.k^2-49.k^2=-360\)

\(-40.k^2=-360\)

\(k^2=9\)

\(k=\pm3\)

+ Nếu \(k=3\)\(\Rightarrow\hept{\begin{cases}x=-3.3=-9\\y=7.3=21\end{cases}}\)

+ Nếu \(k=-3\)\(\Rightarrow\hept{\begin{cases}x=-3.\left(-3\right)=9\\y=7.\left(-3\right)=-21\end{cases}}\)

\(\left|4-2,5x\right|+20\%=\frac{2}{5}\)

\(\left|4-2,5x\right|+\frac{1}{5}=\frac{2}{5}\)

\(\left|4-2,5x\right|=\frac{1}{5}\)

=>\(\orbr{\begin{cases}4-2,5x=\frac{1}{5}\\4-2,5x=\frac{-1}{5}\end{cases}}\)=>\(\orbr{\begin{cases}2,5x=\frac{19}{5}\\2,5x=\frac{21}{5}\end{cases}}\)=>\(\orbr{\begin{cases}x=\frac{38}{25}\\x=\frac{42}{25}\end{cases}}\)