a, \(\left|x\right|=-1,2\)
\(\Rightarrow x\in\varnothing\)
b, \(\left|x\right|+1,5=3,7\)
\(\left|x\right|=3,7-1,5\)
\(\left|x\right|=2,2\)
\(\Rightarrow x\in\left\{-2,2;2,2\right\}\)
c, \(\left|x+\frac{1}{3}\right|-4=-1\)
\(\left|x+\frac{1}{3}\right|=-1+4\)
\(\left|x+\frac{1}{3}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3-\frac{1}{3}\\x=-3-\frac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=-\frac{10}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{8}{3};-\frac{10}{3}\right\}\)
Câu 1 :
\(\left|x\right|=-1,2\)
\(\Rightarrow x\in\left\{\varnothing\right\}\)
Câu 2 :
\(\left|x\right|+1,5=3,7\)
\(\left|x\right|=2,2\)
\(\Rightarrow x=2,2\) hoặc \(x=-2,2\)
Câu 3 :
\(\left|x+\frac{1}{3}\right|-4=-1\)
\(\left|x+\frac{1}{3}\right|=3\)
TH 1:
\(x+\frac{1}{3}=3\)
\(x=\frac{8}{3}\)
TH 2:
\(x+\frac{1}{3}=-3\)
\(x=-\frac{10}{3}\)
Vậy \(x=\frac{8}{3}\) và \(x=-\frac{10}{3}\)
\(\left|x\right|=-1,2\)
Ta luôn có \(\left|x\right|\ge0\) \(\forall x.\)
\(\Rightarrow\left|x\right|>-1,2\)
\(\Rightarrow\left|x\right|\ne-1,2\)
Vậy \(x\in\varnothing.\)
b) \(\left|x\right|+1,5=3,7\)
\(\Rightarrow\left|x\right|=3,7-1,5\)
\(\Rightarrow\left|x\right|=2,2\)
\(\Rightarrow\left[{}\begin{matrix}x=2,2\\x=-2,2\end{matrix}\right.\)
Vậy \(x\in\left\{2,2;-2,2\right\}.\)
c) \(\left|x+\frac{1}{3}\right|-4=-1\)
\(\Rightarrow\left|x+\frac{1}{3}\right|=\left(-1\right)+4\)
\(\Rightarrow\left|x+\frac{1}{3}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3-\frac{1}{3}\\x=\left(-3\right)-\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=-\frac{10}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{8}{3};-\frac{10}{3}\right\}.\)
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