3х²<=18х

To solve the inequality \( 3x^2 \leq 18x \), we can first rearrange it into a standard form:

1. Subtract \( 18x \) from both sides: \[ 3x^2 - 18x \leq 0 \]

2. Factor the left-hand side: \[ 3x(x - 6) \leq 0 \]

Now, we will find the critical points by setting the factors equal to zero:

1. \( 3x = 0 \) gives \( x = 0 \)
2. \( x - 6 = 0 \) gives \( x = 6 \)

This means the critical points are \( x = 0 \) and \( x = 6 \).

Next, we test the intervals defined by these critical points:

1. Interval \( (-\infty, 0) \)
2. Interval \( (0, 6) \)
3. Interval \( (6, +\infty) \)

Testing intervals:

• For the interval \( (-\infty, 0) \), choose \( x = -1 \): \[ 3(-1)(-1 - 6) = 3(-1)(-7) = 21 \quad \text{(positive)} \]

• For the interval \( (0, 6) \), choose \( x = 1 \): \[ 3(1)(1 - 6) = 3(1)(-5) = -15 \quad \text{(negative)} \]

• For the interval \( (6, +\infty) \), choose \( x = 7 \): \[ 3(7)(7 - 6) = 3(7)(1) = 21 \quad \text{(positive)} \]

Summary of signs:

• \( (-\infty, 0) \): positive
• \( (0, 6) \): negative
• \( (6, +\infty) \): positive

The inequality \( 3x(x - 6) \leq 0 \) is satisfied in the interval where the expression is negative, which is \( (0, 6) \), and also at the points where the expression is zero, which are \( x = 0 \) and \( x = 6 \).

Final solution: \[ x \in [0, 6] \]