Question

-(x+1)²-(x-1)²+12≥0
-x²+5x≤0

inecuaciones de segundo grado...
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Answer 1

Explicación paso a paso:

Primer problema:

-(x + 1)² - (x - 1)² + 12 ≥ 0

-(x² + 2x + 1) - (x² - 2x + 1) + 12 ≥ 0

-x² - 2x - 1 - x² + 2x - 1 + 12 ≥ 0

-2x + 10 ≥ 0

-2x² ≥ -10

|x| ≤ √5

x ≤ √5, x ≥ 0

-x ≤ √5, x < 0

x ∈ [0, √5]

x ≥ - √5, x < 0

x ∈ [0, √5]

x ∈ [-√5, 0》

x ∈ [-√5, √5]

Segundo problema:

-x² + 5x ≤ 0

-x × (x - 5) ≤ 0

-x × (x - 5) ≥ 0

{x ≥ 0

{x - 5 ≥ 0

{x ≤ 0

{x - 5 ≤ 0

{x ≥ 0

{x ≥ 5

{x ≤ 0

{x ≤ 5

x ∈ [5, + ∞》

x ∈ 《-∞, 0]

x ∈ 《-∞, 0] ∪ [5, + ∞》

Besitos OvO