Đáp án + giải thích bước giải :
`a) |x + 3| - |2x + 5| = 0`
`⇔ |x + 3| = 0 + |2x + 5|`
`⇔ |x + 3| = |2x + 5|`
`⇔` \(\left[ \begin{array}{l}x+3=2x+5\\x+3=-2x-5\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x-2x=-3+5\\x+2x=-3-5\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}-x=2\\3x=-8\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-2\\x=\dfrac{-8}{3}\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=-2\\x=\dfrac{-8}{3}\end{array} \right.\)
`2) |x - 4| + |y - 6| + |x-y-z|=0`
Ta có : \(\left\{ \begin{array}{l}|x-4|≥0∀x\\|y-6|≥0∀y\\|x-y-z|≥0∀z\end{array} \right.\)
`⇔ |x - 4| + |y - 6| + |x-y-z|≥0∀x,y,z`
Dấu "`=`" xảy ra khi :
`⇔` \(\left\{ \begin{array}{l}x-4=0\\y-6=0\\x-y-z=0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x=4\\y=6\\4-6-z=0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x=4\\y=6\\z=2\end{array} \right.\)
Vậy \(\left\{ \begin{array}{l}x=4\\y=6\\z=2\end{array} \right.\)
