1) (3x²+4x)/(9-x)>0 2) log0.25 (3x-5)>-3 3) 2cos(x/2)+1=0

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вопрос

1) To find the solution to the inequality (3x²+4x)/(9-x)>0, we first need to determine the critical points where the expression is equal to zero or undefined. The critical points are x=0 and x=9.

Next, we need to test the intervals created by the critical points. Let's test the intervals (-∞, 0), (0, 9), and (9, +∞) by choosing test points within each interval. For example, if we choose x=-1 for the interval (-∞, 0), we get (3(-1)²+4(-1))/(9-(-1)) = (-3-4)/10 = -7/10, which is negative. This means that the expression is negative in the interval (-∞, 0).

Similarly, for the interval (0, 9), choosing x=1 gives us (3(1)²+4(1))/(9-1) = (3+4)/8 = 7/8, which is positive. This means that the expression is positive in the interval (0, 9).

Therefore, the solution to the inequality is x∈(0, 9).

2) To solve the logarithmic inequality log0.25 (3x-5)>-3, we need to rewrite it in exponential form. The inequality becomes 0.25^(-3) < 3x-5.

Simplifying the left side, we get 4^3 < 3x-5, which is equivalent to 64 < 3x-5.

Adding 5 to both sides, we get 69 < 3x.

Dividing by 3, we find x > 23. So the solution to the inequality is x > 23.

3) To solve the equation 2cos(x/2)+1=0, we need to isolate the cosine term first. Subtracting 1 from both sides, we get 2cos(x/2) = -1.

Dividing by 2, we get cos(x/2) = -1/2.

This means that the angle x/2 must be in the second or third quadrant where cosine is negative and has a value of -1/2. Two possible solutions are x = 150 degrees or x = 210 degrees.