Question

Решите рациональные неравенства 11-(x+1)^2≥x (2x-8)^2-4x(2x-8)≥0 x(x+5)-2>4x (1/3)x^2+3x+6<0 x>(x^2/2)-4x+5(1/2)

Answer 1

$1)\; \; 11-(x+1)^2\geq x\\\\11-x^2-2x-1\geq x\\\\x^2+3x-10\leq 0\; ,\; \; x_1=-5\; ,\; x_2=2\; (teorema\; Vieta)\\\\znaki:\quad +++[-5\, ]---[\, 2\, ]+++\\\\x\in [-5,2\, ]\\\\2)\; \; (2x-8)^2-4x(2x-8)\geq 0\\\\4x^2-32x+64-8x^2+32x\geq 0\\\\-4x^2+64\geq 0\; |:(-4)\; \; \to \; \; x^2-16\leq 0\\\\(x-4)(x+4)\leq 0\; ,\qquad +++[-4\, ]---[\, 4\, ]+++\\\\x\in [-4,4\, ]\\\\3)\; \; x(x+5)-2>4x\\\\x^2+5x-4x-2>0\; \; \to \\\\x^2+x-2>0\; ,\; \; x_1=-2\; ,\; x_2=1\; (teorema\; Vieta)\\\\(x+2)(x-1)>0\quad +++(-2)---(1)+++\\\\x\in (-\infty ,-2)\cup (1,+\infty )$

$4)\; \; \frac{1}{3}x^2+3x+6$

Answer 2

11-(x+1)²≥x

11-x²-2x-1≥x

x²+3x-10≤0

x²+3x-10=0 D=49 √D=7

x₁=2 x₂=-5

(x-2)(x+5)≤0

-∞____+____-5___-____2____+____+∞ ⇒

Ответ: x∈[-5;2]

(2x-8)²-4x*(2x-8)≥0

(2x-8)(2x-8-4x)≥0

(2x-8)(-2x-8)≥0

-(2x-8)*(2x+8)≥0 |÷(-1)

4x²-64≤0 |÷4

x²-16≤0

(x-4)(x+4)≤0

-∞____+____-4____-____4____+____+∞ ⇒

Ответ: x∈[-4;4].

x*(x+5)-2>4x

x²+5x-2-4x>0

x²+x-2>0

x²+x-2=0 D=9 √D=3

x₁=1 x₂=-2 ⇒

(x-1)(x+2)>0

-∞____+____-2____-____1____+____+∞ ⇒

Ответ: x∈(-∞-2)U(1;+∞).

(1/3)*x²+3x+6<0 |×3

x²+9x+18<0

x²+9x+18=0 D=9 √D=3

x₁=-3 x₂=-6 ⇒

(x+3)(x+6)<0

-∞_____+_____-6_____-_____-3____+_____+∞ ⇒

Ответ: x∈(-6;-3).

x>(x²/2)-4x+5¹/₂

x>(x²/2)-4x+11/2 |×2

2x>x²-8x+11

x²-10x+11<0

x²-10x+11=0 D=56 √D=√56

x₁=5-√14 x₂=5+√14

-∞____+_____5-√14____-_____5+√14____+____+∞ ⇒

Ответ: x∈(5-√14;5+√14).

Answer 3

To solve rational inequalities, we need to find the intervals for which the given expressions are true. Let's solve each inequality step by step:

1. 11 - (x + 1)^2 ≥ x First, expand the squared term: 11 - (x + 1)(x + 1) ≥ x

Simplify further: 11 - (x^2 + 2x + 1) ≥ x 11 - x^2 - 2x - 1 ≥ x

Combine like terms: 10 - x^2 - 2x ≥ x

Move all terms to one side of the inequality: 10 - x^2 - 2x - x ≥ 0 10 - x^2 - 3x ≥ 0

Now, set each factor equal to zero and find the critical points: 10 - x^2 - 3x = 0 -x^2 - 3x + 10 = 0

Using the quadratic formula, the critical points are: x = (-(-3) ± √((-3)^2 - 4*(-1)10)) / (2(-1)) x = (3 ± √(9 + 40)) / (-2) x = (3 ± √49) / (-2) x = (3 ± 7) / (-2)

So, the critical points are x = 5 and x = -2.

Now, we create a sign chart and test the intervals:

markdownCopy code
Interval        |  x^2 + 3x - 10  |  10 - x^2 - 3x
----------------------------------------------
x < -2          |     +           |      +
-2 < x < 5      |     -           |      +
x > 5           |     +           |      -

The solution is x < -2 or x > 5.

2. (2x - 8)^2 - 4x(2x - 8) ≥ 0 First, factor the quadratic expression: (2x - 8)(2x - 8) - 4x(2x - 8) ≥ 0 (2x - 8)(2x - 8 - 4x) ≥ 0 (2x - 8)(-2x - 8) ≥ 0 -(2x - 8)^2 ≥ 0

Since the square of a real number is non-negative, the inequality is true for all real values of x. Thus, the solution is (-∞, +∞).

3. x(x + 5) - 2 > 4x First, distribute x into the parenthesis: x^2 + 5x - 2 > 4x

Move all terms to one side of the inequality: x^2 + 5x - 4x - 2 > 0 x^2 + x - 2 > 0

Now, set the quadratic expression equal to zero and find the critical points: x^2 + x - 2 = 0

Using the quadratic formula, the critical points are: x = (-1 ± √(1 + 8)) / 2 x = (-1 ± √9) / 2 x = (-1 ± 3) / 2

So, the critical points are x = 1 and x = -2.

Now, we create a sign chart and test the intervals:

markdownCopy code
Interval        |  x^2 + x - 2
--------------------------------
x < -2          |     -      
-2 < x < 1      |     +
x > 1           |     +

The solution is x < -2 or x > 1.

4. (1/3)x^2 + 3x + 6 < 0 To solve this inequality, let's first multiply both sides by 3 to clear the fraction: x^2 + 9x + 18 < 0

Now, set the quadratic expression equal to zero and find the critical points: x^2 + 9x + 18 = 0

Using the quadratic formula, the critical points are: x = (-9 ± √(9^2 - 4118)) / 2 x = (-9 ± √(81 - 72)) / 2 x = (-9 ± √9) / 2 x = (-9 ± 3) / 2

So, the critical points are x = -6 and x = -3.

Now, we create a sign chart and test the intervals:

markdownCopy code
Interval        |  x^2 + 9x + 18
--------------------------------
x < -6          |     +      
-6 < x < -3     |     -
x > -3          |     +

The solution is -6 < x < -3.

5. x > (x^2/2) - 4x + 5/2 To solve this inequality, let's first move all terms to one side of the inequality: x - (x^2/2) + 4x - 5/2 > 0 (2x - x^2 + 8x - 5)/2 > 0 (10x - x^2 - 5)/2 > 0 (5 - x^2 + 10x)/2 > 0 (5 - x^2 + 10x) > 0 * 2 (Note: Multiplying both sides by a positive number doesn't change the inequality direction)

Now, set the quadratic expression equal to zero and find the critical points: x^2 - 10x + 5 = 0

Using the quadratic formula, the critical points are: x = (10 ± √(10^2 - 415)) / 2 x = (10 ± √(100 - 20)) / 2 x = (10 ± √80) / 2 x = (10 ± 4√5) / 2 x = 5 ± 2√5

So, the critical points are x = 5 + 2√5 and x = 5 - 2√5.

Now, we create a sign chart and test the intervals:

markdownCopy code
Interval        |  5 - x^2 + 10x
--------------------------------
x < 5 - 2√5     |     -      
5 - 2√5 < x < 5 + 2√5 |     +
x > 5 + 2√5     |     -

The solution is x < 5 - 2√5 or x > 5 + 2√5.

To summarize the solutions:

1. x < -2 or x > 5
2. (-∞, +∞)
3. x < -2 or x > 1
4. -6 < x < -3
5. x < 5 - 2√5 or x > 5 + 2√5