Question

РЕШИТЕ ПОЖАЛУЙСТА 1.) x²>812.) 3x²-8x+5<03.) 3x²-8x+5<04.) (x+7)(x-12)(x-9)≥0​

Answer 1

Ответ:

Объяснение:

1). x²>81

x²-9²>0

(x-9)(x+9)>0

Допустим (x-9)(x+9)=0

x-9=0; x₁=9

x+9=0; x₂=-9

Для определения знака функции возьмём пробную точку на промежутке (-9; 9), например, 0.

0² ∨ 81; 0² ∨ 9²; 0<9

Неравенство не выполняется, следовательно, знак на этом интервале будет минус.

        +                   -                       +

----------------°--------------------°--------------->x

                -9                      9

Ответ: x∈(-∞; -9)∪(9; +∞).

2). 3x²-8x+5<0

Допустим 3x²-8x+5=0

D=64-60=4

x₁=(8-2)/6=6/6=1

x₂=(8+2)/6=10/6=5/3=1 2/3

Для определения знака функции возьмём пробную точку на промежутке (-∞; 1), например, 0.

3·0²-8·0+5 ∨ 0; 5>0

Неравенство не выполняется, следовательно, знак на этом интервале будет минус.

       -                   +                   -

-----------------°-----------------°----------------->x

                   1                  1 2/3

Ответ: x∈(1; 1 2/3).

3). Чтобы не повторялось неравенство, поменяем знак.

3x²-8x+5>0

Допустим 3x²-8x+5=0; D=4; x₁=1; x₂=1 2/3

Для определения знака функции возьмём пробную точку на промежутке (-∞; 1), например, 0.

3·0²-8·0+5 ∨ 0; 5>0

Неравенство выполняется, следовательно, знак на этом интервале будет плюс.

         +                       -                     +

--------------------°-------------------°------------------>x

                      1                     1 2/3

Ответ: x∈(-∞; 1)∪(1 2/3; +∞).

4). (x+7)(x-12)(x-9)≥0

Допустим (x+7)(x-12)(x-9)=0

x+7=0; x₁=-7

x-12=0; x₂=12

x-9=0; x₃=9

Для определения знака функции возьмём пробную точку на промежутке [9; 12], например, 10.

(10+7)(10-12)(10-9) ∨ 0

17·(-2)·1 ∨ 0

-34<0

Неравенство не выполняется, следовательно, знак на этом интервале будет минус.

            -                           +                          -                         +

-------------------------.-----------------------.-------------------------.----------------->x

                           -7                          9                           12

Ответ: x∈[-7; 9]∪[12; +∞).

Answer 2

1) x² > 81: To solve this inequality, we need to find the values of x that satisfy the given condition. Let's break it down step by step:

First, let's find the square root of both sides of the inequality: √(x²) > √81

This simplifies to: |x| > 9

Now, we have two cases to consider: 1a) x > 9 In this case, all values of x greater than 9 will satisfy the inequality.

1b) x < -9 In this case, all values of x less than -9 will satisfy the inequality.

Therefore, the solution to the inequality x² > 81 is: x < -9 or x > 9

2) 3x² - 8x + 5 < 0: Let's solve this quadratic inequality step by step:

First, let's find the roots of the quadratic equation 3x² - 8x + 5 = 0. We can do this by factoring or using the quadratic formula. However, since we are interested in the sign of the expression, we can use the factored form directly.

The factored form of the quadratic equation is: 3x² - 8x + 5 = (x - 1)(3x - 5) < 0

Next, we need to find the critical points, where the expression changes sign. The critical points occur when one of the factors becomes zero: x - 1 = 0 => x = 1 3x - 5 = 0 => x = 5/3

Now, let's analyze the sign of the expression in each interval: Interval (-∞, 1): Pick a value less than 1, like x = 0. Substitute it into the expression: (0 - 1)(3(0) - 5) = (-1)(-5) = 5, which is positive.

Interval (1, 5/3): Pick a value between 1 and 5/3, like x = 2. Substitute it into the expression: (2 - 1)(3(2) - 5) = (1)(1) = 1, which is positive.

Interval (5/3, +∞): Pick a value greater than 5/3, like x = 2. Substitute it into the expression: (3 - 1)(3(3) - 5) = (2)(4) = 8, which is positive.

Since the expression is positive in all intervals, there are no values of x that satisfy the inequality 3x² - 8x + 5 < 0.

3) 3x² - 8x + 5 ≤ 0: To solve this quadratic inequality, we follow a similar process as in the previous question:

First, let's find the roots of the quadratic equation 3x² - 8x + 5 = 0: 3x² - 8x + 5 = (x - 1)(3x - 5) ≤ 0

The critical points occur when one of the factors becomes zero: x - 1 = 0 => x = 1 3x - 5 = 0 => x = 5/3

Now, let's analyze the sign of the expression in each interval: Interval (-∞, 1): Pick a value less than 1, like x = 0. Substitute it into the expression: (0 - 1)(3(0) - 5) = (-1)(-5) = 5, which is positive.

Interval (1, 5/3): Pick a value between 1 and 5/3, like x = 2. Substitute it into the expression: (2 - 1)(3(2) - 5) = (1)(1) = 1, which is positive.

Interval (5/3, +∞): Pick a value greater than 5/3, like x = 2. Substitute it into the expression: (3 - 1)(3(3) - 5) = (2)(4) = 8, which is positive.

Since the expression is positive in all intervals, there are no values of x that satisfy the inequality 3x² - 8x + 5 ≤ 0.

4) (x + 7)(x - 12)(x - 9) ≥ 0: To solve this inequality, we need to find the values of x that make the expression non-negative (greater than or equal to zero). Let's break it down step by step:

We have three critical points where the expression changes sign: x = -7, x = 12, and x = 9. These points divide the number line into four intervals:

Interval (-∞, -7): Pick a value less than -7, like x = -8. Substitute it into the expression: (-8 + 7)(-8 - 12)(-8 - 9) = (-1)(-20)(-17) = -340, which is negative.

Interval (-7, 9): Pick a value between -7 and 9, like x = 0. Substitute it into the expression: (0 + 7)(0 - 12)(0 - 9) = (7)(-12)(-9) = 756, which is positive.

Interval (9, 12): Pick a value between 9 and 12, like x = 10. Substitute it into the expression: (10 + 7)(10 - 12)(10 - 9) = (17)(-2)(1) = -34, which is negative.

Interval (12, +∞): Pick a value greater than 12, like x = 13. Substitute it into the expression: (13 + 7)(13 - 12)(13 - 9) = (20)(1)(4) = 80, which is positive.

From this analysis, we can see that the expression is negative in the intervals (-∞, -7) and (9, 12), and positive in the intervals (-7, 9) and (12, +∞).

Therefore, the solution to the inequality (x + 7)(x - 12)(x - 9) ≥ 0 is: x ≤ -7 or 9 ≤ x ≤ 12