Question

√(х+7)≤2 решите неравенства

Answer 1

√(х+7) ≤ 2

ОДЗ: х ≥ -7

х+7 ≤ 4

х ≤ 4-7

х ≤ -3

Объединяем:

х ≤ -3

х ≥ -7

=> х Є [-7; -3]

Ответ: х Є [-7; -3]

Answer 2

$\sqrt{x+7} \leq 2$

Возводим обе стороны в квадрат, чтобы избавиться от корня.

x+7 ≤ 4

x+7-4 ≤ 0

x+3 ≤ 0

x = -3, наша точка.

//////////////////////////*----------------------------------------------------->

-∞                         -3                                                           +∞

x ∈ (-∞;-3].

Answer 3

To solve the inequality √(x + 7) ≤ 2, follow these steps:

Step 1: Subtract 2 from both sides to isolate the square root expression: √(x + 7) - 2 ≤ 0

Step 2: Square both sides of the inequality to remove the square root (remember that when squaring both sides of an inequality, the direction of the inequality may change, so we will consider both cases): (√(x + 7) - 2)² ≤ 0

Step 3: Simplify the left-hand side: (x + 7) - 4√(x + 7) + 4 ≤ 0

Step 4: Move all terms to one side of the inequality: x + 7 - 4√(x + 7) + 4 ≤ 0

Step 5: Combine like terms: x + 11 - 4√(x + 7) ≤ 0

Now, we have two cases to consider:

Case 1: √(x + 7) - 2 ≤ 0 In this case, the square root is less than or equal to 2:

√(x + 7) ≤ 2

To solve this, isolate the square root:

x + 7 ≤ 2² x + 7 ≤ 4

Now, subtract 7 from both sides:

x ≤ 4 - 7 x ≤ -3

Case 2: √(x + 7) - 2 ≥ 0 In this case, the square root is greater than or equal to 2:

√(x + 7) ≥ 2

To solve this, square both sides (remember that when squaring both sides, the inequality direction remains the same):

x + 7 ≥ 2² x + 7 ≥ 4

Now, subtract 7 from both sides:

x ≥ 4 - 7 x ≥ -3

So the solution to the inequality is -3 ≤ x ≤ -3, meaning x is in the closed interval [-3, -3].