Tìm x, biết |3-2x|-x+2=0

a, \(\left|3-2x\right|-x+2=0\)

+, Xét \(x\le\dfrac{3}{2}\Rightarrow3-2x\ge0\Rightarrow\left|3-2x\right|=3-2x\) ta có:

\(3-2x-x+2=0\)

\(\Rightarrow-3x=-2-3\Rightarrow x=\dfrac{5}{3}\)(laoị vì không thoả mãn điều kiện \(x\le\dfrac{3}{2}\))

+,Xét \(x>\dfrac{3}{2}\Rightarrow3-2x< 0\Rightarrow\left|3-2x\right|=2x-3\) ta có:

\(2x-3-x+2=0\)

\(\Rightarrow x=-2+3\Rightarrow x=1\)(loại vì không thoả mãn điều kiện \(x>\dfrac{3}{2}\))
Vậy \(x\in\varnothing\)

b, \(3-2\left|x+4\right|=1-3\)

\(\Rightarrow2\left|x+4\right|=3-1+3\)

\(\Rightarrow\left|x+4\right|=\dfrac{5}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x+4=\dfrac{5}{2}\\x+4=-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\x=-\dfrac{13}{2}\end{matrix}\right.\)

Vậy...............

c, \(\left|x-3\right|^2=1^2-2^2+\left(-3\right)^2+3\)

\(\Rightarrow\left(x-3\right)^2=1-4+9+3\)(do \(\left|A\left(x\right)\right|^2=\left[A\left(x\right)\right]^2\))

\(\Rightarrow\left\{{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\x=0\end{matrix}\right.\)

Vậy...............

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