Question

X(x+3)+2x<(или ровно)(х+1)^2+1​

Answer 1

Ответ:

Пошаговое объяснение:

x^2+3x+2x<=x^2+2x+2

3x<=2

x<=2/3

(-∞;2/3]

Answer 2

Ответ:

(-∞; 2/3].

Пошаговое объяснение:

х(x+3)+2x ≤ (х+1)^2+1​

х ^2 + 3х + 2х ≤ х ^2 + 2х + 1 + 1

х ^2 + 5х ≤ х ^2 + 2х + 2

х ^2 + 5х - х ^2 - 2х ≤ 2

3х ≤ 2

х ≤ 2/3

х∈ (- ∞; 2/3]

Ответ: (-∞; 2/3].

Answer 3

To solve the inequality $X(x+3)+2x \leq (x+1)^2+1$, we'll first simplify and then solve for the variable $x$. The process involves expanding and simplifying both sides of the inequality.

Let's start step by step:

1. Expand both sides of the inequality: $X(x+3) + 2x \leq (x+1)^2 + 1$

   Expand the right-hand side: $X(x) + X(3) + 2x \leq x^2 + 2x + 1 + 1$

2. Simplify: $Xx + 3X + 2x \leq x^2 + 2x + 2$

3. Move all terms to the left-hand side to set the inequality to zero: $Xx + 3X + 2x - x^2 - 2x - 2 \leq 0$

4. Combine like terms: $Xx + 2x - x^2 + 3X - 2 \leq 0$

5. Arrange the terms in descending order of $x$: $-x^2 + (X + 2)x + (3X - 2) \leq 0$

The inequality is now in the form $ax^2 + bx + c \leq 0$, where $a = -1$, $b = X + 2$, and $c = 3X - 2$.

To find the solution for $x$, we need to determine the values of $X$ for which the inequality is true. We can do this by analyzing the sign of the quadratic expression $ax^2 + bx + c$ based on the value of $a$ and the discriminant $\Delta = b^2 - 4ac$.

For the inequality $ax^2 + bx + c \leq 0$ to hold true, the quadratic must have real roots (or a repeated real root) and the sign of $a$ must be negative.

1. Discriminant: $\Delta = (X + 2)^2 - 4(-1)(3X - 2) = (X^2 + 4X + 4) + 12X - 8 = X^2 + 16X - 4$

2. Find the critical points (where the discriminant is zero): $\Delta = 0 \implies X^2 + 16X - 4 = 0$

To solve this quadratic equation, we can use the quadratic formula: $X = \frac{-b \pm \sqrt{\Delta}}{2a}$ where $a = 1$, $b = 16$, and $c = -4$.

3. Calculate the roots: $X = \frac{-16 \pm \sqrt{(16)^2 - 4(1)(-4)}}{2(1)}$ $X = \frac{-16 \pm \sqrt{272}}{2}$ $X = \frac{-16 \pm 4\sqrt{17}}{2}$ $X = -8 \pm 2\sqrt{17}$

So, the critical points are $X = -8 + 2\sqrt{17}$ and $X = -8 - 2\sqrt{17}$.

4. Analyze the sign of $a$ for different intervals:

   a) 0 0