Question

1) 4^x-2^x+5-68≥0 2) 21^x-9*7^x-3^x+9<0 3) 2*4^-x - 33*2^-x+16≤0 4)2^(х в квадрате) ≤4*2^х Помогите,пожалуйста

Answer 1

3) 2*(2^2)^-x-33*2^-x+16 ≤0
Замена  2^-x=y
2y^2-33y+16 ≤0
Через дискриминант ищем корни.
у=16, у=1/2
Возвращаемся в замену.
2^-x=16, x=-4
2^-x=1/2, x=1

Answer 2

1. 4^x - 2^x + 5 - 68 ≥ 0

Simplifying the left-hand side, we get:

4^x - 2^x - 63 ≥ 0

Now, let's substitute y = 2^x. Then we have:

y^2 - y - 63 ≥ 0

Solving for y, we get:

y ≤ -8 or y ≥ 9

Substituting back, we get:

2^x ≤ -8 or 2^x ≥ 9

The first inequality is not possible, since 2^x is always positive. Therefore, we have:

2^x ≥ 9

Taking the logarithm base 2 of both sides, we get:

x ≥ log2(9)

So the solution is:

x ≥ 3.16993

2. 21^x - 9*7^x - 3^x + 9 < 0

We can notice that 21 = 3*7, so we can substitute y = 7^x. Then we have:

3y^2 - 9y - y + 9 < 0

Simplifying, we get:

3y(y - 3) - (y - 3) < 0

Factoring, we get:

(y - 3)(3y - 1) < 0

Therefore, we have:

y < 1/3 or y > 3

Substituting back, we get:

7^x < 1/3 or 7^x > 3

Taking the logarithm base 7 of both sides, we get:

x < log7(1/3) or x > log7(3)

Using a calculator, we get:

x < -0.82067 or x > 0.77124

So the solution is:

-0.82067 < x < 0.77124

3. 24^-x - 332^-x + 16 ≤ 0

Let's substitute y = 2^-x. Then we have:

2/y^2 - 33/y + 16 ≤ 0

Multiplying both sides by y^2, we get:

2 - 33y + 16y^2 ≤ 0

Solving for y, we get:

y ≤ 1/2 or y ≥ 8/16

Substituting back, we get:

2^-x ≤ 1/2 or 2^-x ≥ 1/2

The second inequality is always true, so we can ignore it. The first inequality is equivalent to:

x ≥ log2(1/2)

Simplifying, we get:

x ≥ -1

So the solution is:

x ≥ -1

4. 2^(x^2) ≤ 4*2^x

Dividing both sides by 2^x, we get:

2^(x^2 - x) ≤ 4

Taking the logarithm base 2 of both sides, we get:

x^2 - x ≤ 2

Rearranging, we get:

x^2 - x - 2 ≤ 0

Factoring, we get:

(x - 2)(x + 1) ≤ 0

Therefore, we have:

-1 ≤ x ≤ 2

So the solution is:

-1 ≤ x ≤ 2