Question

Giải phương trình:
a) |2x+2| + 10 = 2x
b) |x - 6| = |3 - 2x|
c) | x^2 - 9| + |2x - 6| = 0

Answer 1

\(a,\left|2x+2\right|+10=2x\)

*TH1 : \(\left|2x+2\right|=2x+2\Leftrightarrow2x+2>0\Leftrightarrow x>-1\)

\(\Rightarrow2x+2+10=2x\)

\(\Leftrightarrow2x-2x=-10-2\)

\(\Leftrightarrow0x=-12\left(vô\cdot lý\right)\)

*TH2 :\(\left|2x+2\right|=-2x-2\Leftrightarrow-2x-2< 0\Leftrightarrow x>-1\)

\(\Rightarrow-2x-2+10=2x\)

\(\Leftrightarrow-2x-2x=-10+2\)

\(\Leftrightarrow-4x=-8\)

\(\Leftrightarrow x=\dfrac{1}{2}\left(nhận\right)\)

Vậy \(S=\left\{\dfrac{1}{2}\right\}\)

\(b,\left|x-6\right|=\left|3-2x\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=3-2x\\x-6=-3+2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Vậy \(S=\left\{-3;3\right\}\)