a, |x+1| +2x \(=\)-2
\(\Leftrightarrow\) |x+1|\(=\)-2-2x (*)
TH1: Nếu x+1\(\ge0\)\(\Leftrightarrow x\ge-1\)thì \(\left|x+1\right|=x+1\)
Thay vào (*) ta có:
\(x+1=-2-2x\)
\(\Leftrightarrow x+2x=-2-1\)
\(\Leftrightarrow3x=-3\)
\(\Leftrightarrow x=-1\)(TMĐK)
TH2: Nếu \(x+1< 0\Leftrightarrow x< -1\Rightarrow\left|x+1\right|=-\left(x+1\right)\)
Thay vào (*), ta có:
\(-x-1=\)\(-2-2x\)
\(\Leftrightarrow-x+2x=-2+1\)
\(\Leftrightarrow x=-1\)(kTMĐK)
Vậy S\(=\){-1}
b, \(x^2-6x+9=1\)
\(\Leftrightarrow x^2-2.3.x+3^2\)
\(\Leftrightarrow\left(x-3\right)^2=1\)
\(\Leftrightarrow\left(x-3\right)^2-1^2=0\)
\(\Leftrightarrow\left(x-3-1\right)\left(x-3+1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Vậy S\(=\){4;2}
a) |x + 1| + 2x = 2
\(\Rightarrow\) |3x + 1| = 2
\(\Rightarrow\) |3x| = 1
\(\Rightarrow\) |x| = 1 : 3 = \(\dfrac{1}{3}\)
\(\Rightarrow\) x = \(\dfrac{1}{3}\) hoặc \(-\dfrac{1}{3}\)