sqrt (2x+3) + sqrt (x-2) = 4

√ (2x + 3) + √ (x - 2) = 4

ОДЗ:

2x + 3 ≥ 0

x - 2 ≥ 0

x ≥ 2

√ (2x + 3) = 4 - √ (x - 2)

2x + 3 = 16 - 8√ (x - 2) + x - 2

2x - x + 3 - 16 + 2 = - 8√ (x - 2)

x - 11 = - 8√ (x - 2)

11 - x = 8√ (x - 2) 11 - x ≥ 0

121 - 22x + x² = 64x - 128 x ≤ 11

x² - 86x + 249 = 0

x₁ + x₂ = 86

x₁x₂ = 249

x₁ = 3; x₂ = 83 - не уд. условию x ≤ 11

Ответ: x = 3.