a,|x-2|=3
=>x-2= 3 hoặc -3
-với x -2 =3 =>x=1
-với x-2 = -3 =>x= -1
Vậy x ∈ {1;-1}
b, |x-3|-7=13
=>|x-3|=13-7=6
=>x-3=6 hoặc -6
-với x-3 =6 => x=9
-với x-3= -6 => x=-3
Vậy x ∈ {9;-3}
c,72-3.|x+1|=9
=>3.|x+1|=72-9=63
=>|x+1|=63:3=21
=>x+1 = 21 hoặc -21
-với x+1 = 21 => x=20
-với x+1 = -21 => x= -22
Vậy x ∈ {20;-21}
d,17-(43-|x|)=45
=>43-|x|=17-45= -28
=>|x|=43-(-28)=71
=> x =71 hoăc -71
Vậy x ∈ {71;-71}
f, 3|x-1|-5=7
=>3|x-1|=7+5=12
=>|x-1|=12:3=4
=>x-1=4 hoặc -4
-với x-1=4 =>x=5
-với x-1=-4=>x=-3
Vậy x ∈ {5;-3}
CHÚC BẠN LÀM BÀI TÔT NHA
a) \(\left|x-2\right|=3\\ \Rightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3+2\\x=-3+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)Vậy \(x\in\left\{5;-1\right\}\)
b) \(\left|x-3\right|-7=13\\ \left|x-3\right|=13+7\\ \left|x-3\right|=20\\ \Rightarrow\left[{}\begin{matrix}x-3=20\\x-3=-20\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=20+3\\x=-20+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=23\\x=-17\end{matrix}\right.\)
Vậy \(x\in\left\{23;-17\right\}\)
c) \(72-3\cdot\left|x+1\right|=9\\ 3\left|x+1\right|=72-9\\ 3\left|x+1\right|=63\\ \Rightarrow\left[{}\begin{matrix}x+1=63\\x+1=-63\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=63-1\\x=-63-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=62\\x=-64\end{matrix}\right.\)
Vậy \(x\in\left\{62;-64\right\}\)
d) \(17-\left(43-\left|x\right|\right)=45\\ 43-\left|x\right|=17-45\\ 43-\left|x\right|=-28\\ \left|x\right|=43-\left(-28\right)\\ \left|x\right|=43+28\\ \left|x\right|=71\\ \Rightarrow x\in\left\{71;-71\right\}\)Vậy \(x\in\left\{71;-71\right\}\)
e) \(\left|x\right|< 3\\ \left|x\right|\in\left\{0;1;2\right\}\\ \Rightarrow x\in\left\{0;\pm1;\pm2\right\}\)Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
f) \(3\left|x-1\right|-5=7\\ 3\left|x-1\right|=7+5\\ 3\left|x-1\right|=12\\ \left|x-1\right|=12:3\\ \left|x-1\right|=4\\ \Rightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{5;-3\right\}\)