\(\left|x-1,5\right|=2\\ \Rightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
Vậy \(x\in\left\{3,5;-0,5\right\}\)
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\(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\\ \Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\\ \Rightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-\frac{5}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};-\frac{5}{4}\right\}\)
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\(\left|x-2\right|=x\left(ĐK:x\ge0\right)\\ \Rightarrow\left[{}\begin{matrix}x-2=x\\x-2=-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-x=2\\x+x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2\left(\text{vô lý}\right)\\2x=2\end{matrix}\right.\\ \Rightarrow x=1\left(tmđk\right)\)
Vậy \(x=1\)
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\(\left|x-3,4\right|+\left|2,6-x\right|=0\\ \Rightarrow\left|x-3,4\right|=-\left|2,6-x\right|\)
Mà \(\left|2,6-x\right|\ge0\forall x\Rightarrow-\left|2,6-x\right|\le0\forall x\)
\(\Rightarrow\left|x-3,4\right|\le0\forall x\left(\text{vô lý}\right)\)
Vậy \(x\in\varnothing\)
a/ \(\left|x-1,5\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2+1,5=3,5\\x=-2+1,5=-0,5\end{matrix}\right.\)
b/ \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=0+\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}-\frac{3}{4}=\frac{2}{4}-\frac{3}{4}=-\frac{1}{4}\\x=-\frac{1}{2}-\frac{3}{4}=\left(-\frac{2}{4}\right)+\left(-\frac{3}{4}\right)=-\frac{5}{4}\end{matrix}\right.\)
c/ \(\left|x-2\right|=x\)
\(\Rightarrow\left[{}\begin{matrix}x-2=x\\x-2=-x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-x=2\\x+x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0=2\left(vô-lý\right)\\2x=2\end{matrix}\right.\)
=> 2x = 2
=> x = 2 : 2 = 1
d/ \(\left|x-3,4\right|+\left|2,6-x\right|=0\)
Ta có: \(\left\{{}\begin{matrix}\left|x-3,4\right|\ge0\\\left|2,6-x\right|\ge0\end{matrix}\right.\)
=> Để \(\left|x-3,4\right|+\left|2,6-x\right|=0\) thì \(\left\{{}\begin{matrix}\left|x-3,4\right|=0\\\left|2,6-x\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-3,4=0\\2,6-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0+3,4=3,4\\x=2,6-0=2,6\end{matrix}\right.\)
