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18.03.2013

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3|x - 1| ≤ x + 3.

: 0 ≤ x ≤ 3.


|x2 + 3x| < x + 4.

: -1 - < x < -2 -2 < x < -1 + .


|2x + 1| ≥ 1.

: x ≤ -1 x ≥ 0.


|x2 - 3x| ≥ x + 5.

: x ≤ -1 x ≥ 5.


|2x2 + x - 1| > |x + 1|.

: x < -1 -1 < x < 0, x > 1.


|x| - 2|x + 1| + 3|x + 2| > 4.

: x < -4 x > 0.


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