Question

3х(х-2)+1менше Або доривнюе (х+1)²​

Answer 1

Всё подробно расписано, ошибок нет

Answer 2

Let's solve this inequality together:

$3x(x-2) + 1 \leq (x+1)^2$

First, expand and simplify both sides:

$3x^2 - 6x + 1 \leq x^2 + 2x + 1$

Now, combine like terms:

$2x^2 - 8x \leq 0$

Factor out the common factor:

$2x(x - 4) \leq 0$

Now, we have two critical points, $x = 0$ and $x = 4$, where the expression equals zero. We can use test points in each interval to determine the sign of the expression. Let's consider three intervals: $x < 0$, $0 < x < 4$, and $x > 4$.

1. For $x < 0$, choose $x = -1$:

   $2(-1)(-1 - 4) \leq 0$ (true)

2. For $0 < x < 4$, choose $x = 2$:

   $2(2)(2 - 4) \leq 0$ (true)

3. For $x > 4$, choose $x = 5$:

   $2(5)(5 - 4) \leq 0$ (false)

The solution is $x \leq 0$ or $0 \leq x \leq 4$.