\(a)\) \(\left|x-3\right|+4=\left|-6\right|\)
\(\Leftrightarrow \left|x-3\right|+4=6\)
\(\Leftrightarrow\left|x-3\right|=6-4\)
\(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
Vậy \(x=1\)
\(b)\) \(7-2\cdot\left|x+1\right|=\left(-3\right)^3\)
\(\Leftrightarrow7-2\cdot\left|x+1\right|=-27\)
\(\Leftrightarrow2\cdot\left|x+1\right|=7-\left(-27\right)\)
\(\Leftrightarrow2\cdot\left|x+1\right|=34\)
\(\Leftrightarrow\left|x+1\right|=34\div2\)
\(\Leftrightarrow\left|x+1\right|=17\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=17\\x+1=-17\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=16\\x=-18\end{cases}}\)
Vậy \(x\in\left\{16;-18\right\}\)
Tìm x :
a, |x-3| + 4 = |-6|
<=> |x-3| + 4 = 6
<=> |x-3| = 6 - 4
<=> |x-3| = 2
<=> x - 3 = 2 hoặc x - 3 = -2
<=> x = 2 + 3 hoặc x = ( -2 ) + 3
<=> x = 5 hoặc x = 1
Vậy x € { 5 ; 1 }
b, 7 - 2 . |x+1| = ( - 3 )^3
<=> 7 - 2 . |x+1| = -27
<=> 2 . |x+1| = 7 + 27
<=> 2 . |x+1| = 34
<=> |x+1| = 34 : 2
<=> |x+1| = 17
<=> x + 1 = 17 hoặc x + 1 = -17
<=> x = 17 - 1 hoặc x = ( -17 ) - 1
<=> x = 16 hoặc x = -18
Vậy x € { 16 ; -18 }
