\(\frac{x}{5}=\frac{y}{-3}\)\(\frac{x^2}{5^2}=\frac{y^2}{-3^2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\frac{x^2}{25}-\frac{y^2}{9}=\frac{1}{16}\)

\(=>\frac{x}{5}=\frac{1}{16}\)

\(=>x.16=5\)

\(=>x=\frac{5}{16}\)

Tương tự ta có \(y=\frac{-3}{16}\)

\(\frac{x}{3}=\frac{y}{4}\)

Áp dụng...ta có:

\(\frac{x}{3}-\frac{y}{4}=\frac{192}{-1}=-192\)

\(x=-192:3=-64\)

\(y=-192:4=-48\)

\(\text{Th1: }\hept{\begin{cases}x-1< 0\\x-2\ge0\end{cases}\Rightarrow\hept{\begin{cases}x< 1\\x\ge2\end{cases}\Rightarrow}x\in\varnothing}\)

\(\text{Th2: }\hept{\begin{cases}x-1>0\\x-2\le0\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x\le2\end{cases}\Rightarrow1< x\le}2}\)

\(\text{Khi đó: }\left|x-1\right|+\left|x-2\right|=4\)

\(\Leftrightarrow x-1+2-x=4\)

\(\Leftrightarrow x-x=4-2+1\)

\(\Leftrightarrow0x=3\left(\text{vô lí}\right)\)

Vậy \(x\in\varnothing\)

\(\text{TH3: }\hept{\begin{cases}x-1\ge0\\x-2\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge1\\x\ge2\end{cases}\Rightarrow}x\ge1}\)

\(\text{Khi đó: }\left|x-1\right|+\left|x-2\right|=4\)

\(\Leftrightarrow x-1+x-2=4\)

\(\Leftrightarrow2x=7\)

\(\Leftrightarrow x=\frac{7}{2}\left(\text{nhận}\right)\)

\(\text{TH4: }\hept{\begin{cases}x-1\le0\\x-2\le0\end{cases}\Rightarrow\hept{\begin{cases}x\le1\\x\le2\end{cases}\Rightarrow}x\le2}\)

\(\text{Khi đó: }\left|x-1\right|+\left|x-2\right|=4\)

\(\Leftrightarrow1-x+2-x=4\)

\(\Leftrightarrow-2x=1\)

\(\Leftrightarrow x=-\frac{1}{2}\left(\text{nhận}\right)\)

Vậy \(x\in\left\{\frac{-1}{2};\frac{7}{2}\right\}\)

\(\orbr{\begin{cases}x-1=4\\x-2=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=4+1\\x=4+2\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}\)

Vậy x\(\in\){5;6}