Question

Решите неравенство:3(x-1)(5x+2)-(5x+6)(3x-1)≤0

Answer 1

(3x-3)(5x+2)-(15$x^{2}$+13x-6)≤0

15$x^{2}$-9x-6-15$x^{2}$-13x+6≤0

-22x≤0

22x≥0

x є [ 0; +∞ )

Answer 2

To solve the inequality, we first need to simplify the expression on the left-hand side (LHS) and then find the intervals where it is less than or equal to zero.

Given inequality: 3(x-1)(5x+2) - (5x+6)(3x-1) ≤ 0

Step 1: Expand the expressions within parentheses:

LHS = 3(x)(5x) + 3(x)(2) - 1(5x)(5x) - 1(5x)(2) - (5x)(3x) + (5x)(1) - 6(3x) + 6(1)

Step 2: Simplify further:

LHS = 15x^2 + 6x - 25x^2 - 2x - 15x + 6

Step 3: Combine like terms:

LHS = -10x^2 - 11x + 6

Now we have the quadratic expression -10x^2 - 11x + 6 on the left-hand side of the inequality.

Step 4: Set the expression equal to zero to find the critical points:

-10x^2 - 11x + 6 = 0

Step 5: Factor the quadratic expression:

-10x^2 - 11x + 6 = -(10x - 3)(x + 2)

Now, we can find the critical points by setting each factor equal to zero:

10x - 3 = 0 ---> x = 3/10

x + 2 = 0 ---> x = -2

Step 6: Plot the critical points on a number line and choose test points from each interval to determine the sign of the expression in each interval.

Number line:

-------------o-------------------o--------------

-2 3/10

Step 7: Test points:

Let's choose test points x = -3 (from the interval x < -2), x = 0 (from the interval -2 < x < 3/10), and x = 1 (from the interval x > 3/10).

Step 8: Evaluate the expression at each test point:

For x = -3: LHS = -10(-3)^2 - 11(-3) + 6 = -90 + 33 + 6 = -51

For x = 0: LHS = -10(0)^2 - 11(0) + 6 = 6

For x = 1: LHS = -10(1)^2 - 11(1) + 6 = -15

Step 9: Determine the sign of the expression in each interval:

In the interval x < -2: LHS is negative (-51 < 0). In the interval -2 < x < 3/10: LHS is positive (6 > 0). In the interval x > 3/10: LHS is negative (-15 < 0).

Step 10: Write the solution:

The solution to the inequality is:

x < -2 or x > 3/10

Please note that we use an open circle at -2 and 3/10 in the number line because the inequality is "less than or equal to," but since we only have "<" in the solution, we use an open circle to indicate that the points -2 and 3/10 are not included in the solution set.