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

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

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

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

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

mình hong biết

4ê4e454dtrc5fvybuvygug66uby7b6ug6b = mấy

Ta có : 

\(\hept{\begin{cases}\left|\frac{17}{5}-x\right|>0\\\left|\frac{x}{2}-2\right|>0\end{cases}}\)

\(M\text{à}:\left|\frac{17}{5}-x\right|+\left|\frac{x}{2}-2\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|\frac{17}{5}-x\right|=0\Rightarrow x=\frac{17}{5}\\\left|\frac{x}{2}-2\right|=0\Rightarrow x=4\end{cases}}\)

hỏi j bn ê