a) |x + 1| + 2x = -2
=> |x + 1| = -2 - 2x
=> |x + 1| = -(2 + 2x)
Vì \(\left|x+1\right|\ge0\forall x\) => \(-\left(2+2x\right)\ge0\forall x\)
=> \(2+2x\le0\)
=> \(2x\le-2\)
=> \(x\le-1\)
+ Với x = -1, ta có:
|-1 + 1| + 2.(-1) = -2
|0| + (-2) = -2
0 + (-2) = -2, luôn đúng
+ Với x < -1 thì |x + 1| = -(x + 1) = -x - 1
Ta có: -x - 1 + 2x = -2
=> x - 1 = -2
=> x = -2 + 1 = -1, không thỏa mãn x < -1
Vậy x = -1
b) x2 - 6x + 9 = 1
=> x2 - 3x - 3x + 9 = 1
=> x.(x - 3) - 3.(x - 3) = 1
=> (x - 3).(x - 3) = 1
=> (x - 3)2 = 1
\(\Rightarrow\left[\begin{array}{nghiempt}x-3=1\\x-3=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=2\end{array}\right.\)
Vậy \(x\in\left\{2;4\right\}\)
