Question

1. Dolegions repisacions (x+3)(x - 10) ≤ (x-2)(x-5)

Answer 1

ответ и решение расположены на фото

Answer 2

To solve the inequality (x+3)(x - 10) ≤ (x-2)(x-5), we can start by expanding both sides:

(x+3)(x - 10) = x^2 - 7x - 30 (x-2)(x-5) = x^2 - 7x + 10

Now we have the inequality:

x^2 - 7x - 30 ≤ x^2 - 7x + 10

Next, we can simplify the inequality by canceling out the common terms on both sides:

-30 ≤ 10

Since -30 is indeed less than or equal to 10, this inequality holds true for all values of x. In other words, the solution to the inequality is the set of all real numbers, or (-∞, +∞).