Question

1. |x|+|x+4|≤12 2. |x+5|-|x-3|>10

Answer 1

Стандартный способ с рассмотрением различных случаев раскрытия модулей давно надоел. Есть очевидный, использующий  геометрический смысл модуля (модуль разности чисел равен расстоянию между ними,  поэтому |x| - это расстояние от x до нуля, |x+4| - расстояние от x до минус четырех.  Ясно что сумма расстояний равна 12, когда x = 4 и x = - 8, а меньше 12 - когда мы находимся слева от 4 и справа от - 8. Во второй задаче подобные рассуждения приводят к тому, что решений нет.)

Но мы пойдем другим путем, который мне подсказал Голубев В.И. своими статьями в газете Математика, а затем своей книгой "Решение сложных и нестандартных задач по математике". Каждый желающий может посмотреть эту книгу - она есть в электронном виде, я же здесь буду применять метод без объяснений.

1) $|x|+|x+4|\le 12\Leftrightarrow\left\{\begin{array}{c} x+(x+4)\le 12\\ x-(x+4)\le 12\\ -x+(x+4)\le 12\\ -x-(x+4)\le 12\end{array}\right.\Leftrightarrow\left\{\begin{array}{c} x\le 4\\ -4\le 12\\ 4\le 12\\ x\ge -8\end{array}\right.\Leftrightarrow x\in [-8;4].$  

2) $|x-3|+10$

$\left\{ \begin{array}{c} \left [ {{x+5>x+7} \atop {-x-5>x+7}}\\ \left [ {{x+5>-x+13} \atop {-x-5>-x+13}} \right. \right. \end{array}\right. \Leftrightarrow\left\{ \begin{array}{c} \left [ {{5>7} \atop {x4} \atop {-5>13}} \right. \right. \end{array}\right. \Leftrightarrow x\in \emptyset$

Answer 2

Let's solve these absolute value inequalities step by step:

1. |x| + |x + 4| ≤ 12

To solve this inequality, we need to consider different cases based on the values of x. We'll consider two cases:

Case 1: x ≥ 0 In this case, |x| = x and |x + 4| = |x - (-4)| = |x - (-4)| = |x - 4|, so the inequality becomes: x + |x - 4| ≤ 12

Now we have two subcases:

Subcase 1.1: x - 4 ≥ 0 (x ≥ 4) In this subcase, |x - 4| = x - 4, so the inequality becomes: x + (x - 4) ≤ 12 2x - 4 ≤ 12

Now, solve for x: 2x ≤ 16 x ≤ 8

Subcase 1.2: x - 4 < 0 (x < 4) In this subcase, |x - 4| = -(x - 4) = 4 - x, so the inequality becomes: x + (4 - x) ≤ 12 4 ≤ 12

Since 4 is always less than or equal to 12, there are no restrictions in this subcase.

Case 2: x < 0 In this case, |x| = -x and |x + 4| = |x - (-4)| = |x - (-4)| = |x + 4|, so the inequality becomes: -x + |x + 4| ≤ 12

Now we have two subcases:

Subcase 2.1: x + 4 ≥ 0 (x ≥ -4) In this subcase, |x + 4| = x + 4, so the inequality becomes: -x + (x + 4) ≤ 12 4 ≤ 12

Since 4 is always less than or equal to 12, there are no restrictions in this subcase.

Subcase 2.2: x + 4 < 0 (x < -4) In this subcase, |x + 4| = -(x + 4) = -x - 4, so the inequality becomes: -x - (x + 4) ≤ 12 -2x - 4 ≤ 12

Now, solve for x: -2x ≤ 16 x ≥ -8

Now, we have two conditions for x:

1. For x ≥ 4, -8 ≤ x ≤ 8
2. For x < -4, x < -8

Therefore, the solution for the absolute value inequality is:

x ∈ (-∞, -8] ∪ [-8, 4] ∪ [8, ∞)

2. |x + 5| - |x - 3| > 10

To solve this inequality, consider different cases based on the values of x. We'll consider two cases:

Case 1: x ≥ 3 In this case, |x - 3| = x - 3 and |x + 5| = x + 5, so the inequality becomes: (x + 5) - (x - 3) > 10

Simplify and solve for x: x + 5 - x + 3 > 10 8 > 10

This is not true for any value of x in this case, so there are no solutions in this case.

Case 2: x < 3 In this case, |x - 3| = -(x - 3) = 3 - x and |x + 5| = -(x + 5) = -x - 5, so the inequality becomes: -(x + 5) - (3 - x) > 10

Simplify and solve for x: -x - 5 - 3 + x > 10 -x - 8 > 10

Now, multiply both sides by -1 (and reverse the inequality): x + 8 < -10

Subtract 8 from both sides: x < -10 - 8 x < -18

So, in the second case, the solution is: x < -18

Now, combine the solutions from both cases:

x < -18 (for x < 3)

Therefore, the solution for the absolute value inequality is:

x < -18