a) |x| = 2,5
=>\(\left[\begin{array}{nghiempt}x=2,5\\x=-2,5\end{array}\right.\)
vậy x=2,5 hoặc x=-2,5
b)|x|=-1,2
=>x không có giá trị thỏa mãn |x|\(\ge\) 0
c)|x| + 0,573 = 2
|x| = 2 - 0,573
|x| = 1,427
=>\(\left[\begin{array}{nghiempt}x=1,427\\x=-1,427\end{array}\right.\)
Vậy x = 1,427 hoặc x = -1,427
d) ∣∣x+13∣∣ - 4 = -1
=>|x+\(\frac{1}{3}\)| =-1 + 4
|x+\(\frac{1}{3}\)| = 3
.....................
Vậy x = \(\frac{8}{3}\) hoặc x = \(\frac{-10}{3}\)
a ) \(\left|x\right|=2,5\Rightarrow x=2,5;x=-2,5\)
b ) \(\left|x\right|=-1,2\Rightarrow\left|x\right|\ge0\forall x\Rightarrow x\in\varnothing\)
c ) \(\left|x\right|+0,573=2\)
\(\Rightarrow\)\(\left[\begin{array}{nghiempt}x+0,573=2\\x+0,573=-2\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}2-0,573\\\left(-2\right)-0,573\end{array}\right.\) \(\Rightarrow\)\(\left[\begin{array}{nghiempt}x=1,427\\x=-2,573\end{array}\right.\)
Vậy \(x\in1,427;-2,573\)
d ) \(\left|x+\frac{1}{3}\right|-4=-1\)
\(\Rightarrow\left|x+\frac{1}{3}\right|=3\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=3-\frac{1}{3}\\x=\left(-3\right)-\frac{1}{3}\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{8}{3}\\x=\frac{-10}{3}\end{array}\right.\)
Vậy \(x\in\frac{8}{3};\frac{-10}{3}\)
(Gíup mình với các thần đồng 's hoc24 ơi)