Question

У=√х+4-√х-3 Х+4>0 Х-3>0

Answer 1

{y=√х+4-√х-3                           ОДЗ: {√(x+4)≥0 => x≥-4
{x+4>0 => x>-4                                               {√(x-3)≥0 => x≥3
{x-3>0 => x>3                                                           x≥3
y>√(3+4)-√(3-3)
y>√7
         x∈(3,+∞)
         y∈(√7,+∞)
График

Answer 2

To simplify the expression, we can start by rationalizing the denominator of each square root:

√(x+4) - √(x-3) = (√(x+4) - √(x-3)) * ( ( √(x+4) + √(x-3) ) / (√(x+4) + √(x-3)) )

This simplifies to:

(√(x+4)^2 - √(x-3)^2) / (√(x+4) + √(x-3))

= (x + 4 - (x - 3)) / (√(x+4) + √(x-3))

= 7 / (√(x+4) + √(x-3))

Now we need to check the conditions of the problem:

• x+4 > 0: This means x > -4
• x-3 > 0: This means x > 3

Therefore, the value of x must be greater than 3.

Now we can plug in x=3 to the simplified expression to see if it is defined:

7 / (√(3+4) + √(3-3)) = 7 / (√7 + √0)

The square root of 0 is undefined, so the expression is not defined when x=3.

For any value of x greater than 3, the expression is defined and equals:

7 / (√(x+4) + √(x-3))

For example, when x=5:

7 / (√(5+4) + √(5-3)) = 7 / (√9 + √2) = 7 / (3 + 1.414) ≈ 1.86.