|x-2| ≤ | x+4|
|x-2| - |x+4| ≤ 0
x-2=0 => x=2
x+4=0 => x= -4
x | -oo -4 2 +oo
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x-2 | - - - - - - - - - - - - - - - - - 0 + + + + + + + + + +
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x-4 | - - - - - - - - 0 + + + + + + + + + + + + + + + + +
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I. x∈(-oo; -4)
|x-2| - |x+4| ≤ 0
⇔ -(x-2) + (x+4) ≤ 0
⇔ -x+2+x+4 ≤ 0
⇔ 6 ≤ 0, nu avem solutii in acest interval
II: x∈[-4; 2]
|x-2| - |x+4| ≤ 0
⇔ -(x-2) - (x+4) ≤ 0
⇔ -x+2-x-4 ≤ 0
⇔ -2x -2 ≤0
⇔ -2 ≤ 2x | :2
⇔ -1 ≤ x
cum x∈[-4; 2] ⇒ solutia in acest interval este [-1; 2]
III: x∈(2; +oo)
|x-2| - |x+4| ≤ 0
⇔ (x-2)-(x+4) ≤ 0
⇔ x-2-x-4 ≤ 0
⇔ -6 ≤ 0, deci in intervalul (2; +oo) exista solutii in intervalul (2; +oo)
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x∈[-1; 2] ∪ (2; +oo)
⇒ x∈ [-1; +oo)
