Ответ:

2x^2 + 15x - 12 > 0

2x^2 + 15x - 12 = 0

D = (15)^2 - 4*2*(-12) = 321

x1,2 = -15 ± √321

4

x1 = -15 + √321. x2 = -15 - √321

4 4

(x - -15 + √321) * ( x - -15 - √321) > 0

4 4

{x1 = -15 + √321 > 0, {x1 = -15 + √321 < 0,

(. 4 ( 4

{x2 = -15 - √321 > 0 {x2 = -15 - √321 < 0

4 4

{x1 > -15 + √321, {x1 < -15 + √321,

( 4 ( 4

{x2 > -15 - √321,. {x2 < -15 - √321,

4. 4

Ответ: x є (-∞; -15-√321/4) U (-15+√321/4; +∞)

2. x^2 - 64 < 0

x^2 <64

|x|<8

x<8, x≥0. x>-8, x<0

x є (-8; 8)

3. x^2 > 2,3x

x^2 - 2,3x > 0

x^2 - 23/10x > 0

10x^2 - 23x > 0

x(10x - 23) > 0

{x > 0,

{10x - 23 > 0

{x < 0,

{10x - 23 < 0

{x > 0,

{ x > 23/10

{x < 0,

{x < 23/10

x є (-∞; 0) U ( 23/10; +∞)

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