Question

(1/25)^x-26(1/5)^x+25<=0

Answer 1

Сделаем замену y=(1/5)^x
y^2-26y+25<=0
решим уравнение y^2-26y+25=0
D=26^2-4*25=576, √D=24
y1=1 и y2=25
ответ для нераенства с y
1<=y<=25
1<=(1/5)^x<=25
ln1<=ln(1/5)^x<=ln25
ln1<=x*ln(1/5)<=2*ln5
ln1<=x*ln5^-1<=2*ln5
ln1<=-x*ln5<=2*ln5
-ln1/ln5>=x>=-2*ln5/ln5
-log₅1>=x>=-2
-2<=x<=-log₅1

Answer 2

To solve the inequality (1/25)^x - 26(1/5)^x + 25 ≤ 0, you can use substitution. Let's make a substitution to simplify the equation:

Let y = (1/5)^x. This means that y^2 = (1/25)^x.

Now, substitute y^2 into the inequality:

y^2 - 26y + 25 ≤ 0

This is a quadratic inequality, and we can solve it by factoring. Factor the left side of the inequality:

(y - 25)(y - 1) ≤ 0

Now, you have factored the inequality. To find the solution, we need to consider the signs of the factors on the left-hand side.

1. When (y - 25) ≤ 0 and (y - 1) ≤ 0: This means y ≤ 25 and y ≤ 1.

2. When (y - 25) ≥ 0 and (y - 1) ≥ 0: This means y ≥ 25 and y ≥ 1.

Now, let's combine these conditions:

1. y ≤ 1 and y ≤ 25 (since we take the intersection of both conditions).

Now, remember that y = (1/5)^x, so we need to convert this back into the original variable, x:

(1/5)^x ≤ 1 and (1/5)^x ≤ 25

To simplify further:

1. (1/5)^x ≤ 1
2. (1/5)^x ≤ (1/5)^2

Now, let's solve each part separately:

1. (1/5)^x ≤ 1 This is always true because any positive number raised to any power is always less than or equal to 1.

2. (1/5)^x ≤ (1/5)^2 To solve this part, we can equate the exponents:

   x ≤ 2

So, the solution to the original inequality is:

x ≤ 2

Therefore, the solution to (1/25)^x - 26(1/5)^x + 25 ≤ 0 is x ≤ 2.