Question

(2/3)^x^2*(27/8)^x-1≥2/3

Answer 1

(2/3)ˣ²*(2/3)⁻³ˣ⁺³≥2/3 т.к 0<2/3<1,то
x²-3x+3≤1
x²-3x+2≤0
x²-3x+2=0
D=1
x=1;x=2
x∈[1;2]

Answer 2

To solve the inequality (2/3)^(x^2) * (27/8)^(x-1) ≥ 2/3, we can take the natural logarithm of both sides:

ln[(2/3)^(x^2) * (27/8)^(x-1)] ≥ ln(2/3)

Using the properties of logarithms, we can simplify the left side of the inequality:

ln(2/3)^(x^2) + ln(27/8)^(x-1) ≥ ln(2/3)

Using the power rule of logarithms, we can simplify further:

(x^2)ln(2/3) + (x-1)ln(27/8) ≥ ln(2/3)

Now we can isolate the variable x on one side of the inequality. First, we distribute the ln(27/8):

x^2ln(2/3) + xln(27/8) - ln(27/8) ≥ ln(2/3)

Next, we move all terms containing x to one side:

x^2ln(2/3) + xln(27/8) ≥ ln(2/3) + ln(27/8)

Finally, we factor out x on the left side:

x(xln(2/3) + ln(27/8)) ≥ ln(2/3) + ln(27/8)

Now we can divide both sides by (ln(2/3) + ln(27/8)):

x ≥ (ln(2/3) + ln(27/8))/(ln(2/3) + ln(27/8))

x ≥ 1

Therefore, the solution to the inequality is x ≥ 1.