Question

6) a(a - 2) +1<a²-2a+2​

Answer 1

Ответ:

а - любое число.

Объяснение:

a(a - 2) + 1 < a² - 2a + 2

a(a - 2) + 1 - a² + 2a - 2 < 0

a² - 2a + 1 - a² + 2a - 2 < 0

1 - 2 < 0

-1 < 0 - истинна

Ответ: a ∈ R (или а - любое число).

Answer 2

Ответ:

a= R (пренадлежит любому числу)

Объяснение:

a(a - 2) +1<a²-2a+2

a²-2a+1<a²-2a+2

a²-a²-2a+2a<2-1

0 < 1

a ∈ R

Answer 3

To solve the inequality $a(a - 2) + 1 < a^2 - 2a + 2$, you can start by simplifying and solving it step by step.

Step 1: Expand and simplify both sides of the inequality.

$a(a - 2) + 1 < a^2 - 2a + 2$

First, expand the left side:

$a^2 - 2a + 1 < a^2 - 2a + 2$

Step 2: Subtract $a^2 - 2a + 1$ from both sides of the inequality to isolate the variable on one side.

$a^2 - 2a + 1 - (a^2 - 2a + 1) < a^2 - 2a + 2 - (a^2 - 2a + 1)$

Simplifying further:

$0 < 2 - 1$

$0 < 1$

Step 3: The inequality $0 < 1$ is always true. Any real number you choose for $a$ will satisfy this inequality because zero is always less than one.

So, the solution to the inequality is:

$a$ can be any real number.