Question

Giải hệ phương trình: \(\left\{{}\begin{matrix}\left|x+5\right|+3\left|y-2\right|=12\\\left|x+5\right|-y=10\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}\left|x+5\right|+3\left|y-2\right|=12\\\left|x+5\right|-y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x+5\right|+3\left|y-2\right|=12\\\left|x+5\right|-y+2=12\end{matrix}\right.\)

⇔ | x + 5 | + 3 | y - 2 | = | x + 5 | - y + 2

⇔ 3 | y - 2 | = - ( y - 2 )

⇒ - ( y - 2 ) ≥ 0⇒ | y - 2 | = - ( y - 2 )

⇔ -3 ( y - 2 ) = - ( y - 2 )

⇔ y - 2 = 0 ⇒ y = 2

⇒ | x + 5 | - 2 = 10

⇔| x + 5 | = 12

\(\Leftrightarrow\left[{}\begin{matrix}x+5=12\\-\left(x+5\right)=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-17\end{matrix}\right.\)

Vậy . . . . . . . . .