Question

|x^2+x-2|+|x+4|<=x^2+2x+6

Answer 1

раскрываем модули и получаем систему:

$\\\left[\begin{array}{cccc}\left \{\begin{array}{ccc}x^2+x-2\geq 0\\x+4\geq 0\\x^2+x-2+x+4\leq x^2+2x+6\end{array}\right.\\\left \{\begin{array}{ccc}x^2+x-2$

$\\\left[\begin{array}{cccc}\left \{\begin{array}{ccc}x^2+x-2\geq 0\\x+4\geq 0\\2\leq 6\end{array}\right.\\\left \{\begin{array}{ccc}x^2+x-2$

решаем:

$1)\left \{\begin{array}{ccc}x^2+x-2\geq 0\\x+4\geq 0\\2\leq 6\end{array}\right.\\x^2+x-2\geq 0\\D=1+8=9=3^2\\x_1=\frac{-1+3}{2}=1\\x_2=\frac{-1-3}{2}=-2$

 +         -          +

----[-2]------[1]------>x

$x \in (-\infty;-2]\cup [1;+\infty)$

$x+4\geq 0\\x\geq -4\\x\in [-4;+\infty)\\\\2\leq 6\\x \in R\\\left \{\begin{array}{ccc}x \in (-\infty;-2]\cup [1;+\infty)\\x\in [-4;+\infty)\\x \in R\end{array}\right. \Rightarrow x\in[-4;-2]\cup [1;+\infty)$

$2)\left \{\begin{array}{ccc}x^2+x-2$

 +       -         +

----[-1]-----[0]----->x

$x\in (-\infty;-1]\cup [0;+\infty)$

$\left \{\begin{array}{ccc}x\in (-2;1)\\x\in [-4;+\infty)\\x\in (-\infty;-1]\cup [0;+\infty)\end{array}\right. \Rightarrow x \in(-2;-1]\cup [0;1)$

$3)\left \{\begin{array}{ccc}x^2+x-2< 0\\x+4< 0\\x^2+2x+4\geq 0\end{array}\right.\\x^2+x-2$

$4)\left \{\begin{array}{ccc}x^2+x-2\geq 0\\x+4$

В итоге:

$\left[\begin{array}{cccc}x\in[-4;-2]\cup [1;+\infty)\\x \in(-2;-1]\cup [0;1)\\x \in \varnothing\\ x \in[-6;-4) \end{array}\right. \Rightarrow x \in[-6;-1]\cup [0;+\infty)$

Ответ: $x \in[-6;-1]\cup [0;+\infty)$

Answer 2

To solve the inequality |x^2+x-2|+|x+4| <= x^2+2x+6, we can break it down into cases based on the sign of the expressions inside the absolute value signs.

Case 1: x^2+x-2 is non-negative (x^2+x-2 >= 0) In this case, the absolute value of x^2+x-2 is equal to the expression inside the absolute value signs. Thus, the inequality becomes: (x^2+x-2) + |x+4| <= x^2+2x+6

Simplifying, we have: x^2 + x - 2 + |x+4| <= x^2 + 2x + 6

The x^2 terms cancel out, and we are left with: x - 2 + |x+4| <= 2x + 6

Rearranging terms, we get: |x + 4| - x <= 8

Case 2: x^2+x-2 is negative (x^2+x-2 < 0) In this case, the absolute value of x^2+x-2 is the negation of the expression inside the absolute value signs. Thus, the inequality becomes: -(x^2+x-2) + |x+4| <= x^2+2x+6

Simplifying, we have: -x^2 - x + 2 + |x+4| <= x^2 + 2x + 6

Combining like terms, we get: -2x + 2 + |x+4| <= 2x + 6

Rearranging terms, we have: |x + 4| - 4x <= 4

Now we need to consider the cases separately:

Case 1a: x + 4 is non-negative (x + 4 >= 0) In this case, the absolute value of x + 4 is equal to the expression inside the absolute value signs. Thus, the inequality becomes: x + 4 - x <= 8

The x terms cancel out, and we have: 4 <= 8

This is true for all values of x.

Case 1b: x + 4 is negative (x + 4 < 0) In this case, the absolute value of x + 4 is the negation of the expression inside the absolute value signs. Thus, the inequality becomes: -(x + 4) - x <= 8

Simplifying, we have: -2x - 4 <= 8

Adding 4 to both sides, we get: -2x <= 12

Dividing by -2 (remember to reverse the inequality since we're dividing by a negative number), we have: x >= -6

Case 2a: x + 4 is non-negative (x + 4 >= 0) In this case, the absolute value of x + 4 is equal to the expression inside the absolute value signs. Thus, the inequality becomes: x + 4 - 4x <= 4

Combining like terms, we have: -3x <= 0

Dividing by -3 (remember to reverse the inequality since we're dividing by a negative number), we have: x >= 0

Case 2b: x + 4 is negative (x + 4 < 0) In this case, the absolute value of x + 4 is the negation of the expression inside the absolute value signs. Thus, the inequality becomes: -(x + 4) - 4x <= 4

Simplifying, we have: -5x - 4 <= 4

Adding 4 to both sides, we get: -5x <= 8

Dividing by -5 (remember to reverse the inequality since we're dividing by a negative number), we have: x >= -8/5

Combining the results from the cases:

Case 1a: x >= -6 Case 1b: x >= -6 Case 2a: x >= 0 Case 2b: x >= -8/5

Taking the intersection of these intervals, we find that the solution to the inequality is: x >= -6

Therefore, the inequality |x^2+x-2|+|x+4| <= x^2+2x+6 holds for all values of x greater than or equal to -6.