a, /x-2/ = 5 - 3 = 2
=> x-2 = 2 => x = 2+2=4
=> x-2 = -2 => -2 + 2= 0
b, / 2x - 3 / = /7/
2x - 3 = 7 => 2x = 10 => x = 5
2x - 3 = -7 => 2x = -4 => x = -2
a) |x - 2| + 3 = 5
=> |x - 2| = 5 - 3
=> |x - 2| = 2
=> \(\left\{{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=2+2\\x=\left(-2\right)+2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy x \(\in\) \(\left\{4;0\right\}\).
b) |2x - 3| = |7|
=> \(\left\{{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=7+3=10\\2x=\left(-7\right)+3=-4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=10:2=5\\x=\left(-4\right):2=-2\end{matrix}\right.\)
Vậy x \(\in\) \(\left\{5;-2\right\}\).
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