Sqrt (2x+1) + Sqrt (5x-4) = Sqrt (12x+1)

1. ОДЗ:

{2x + 1 ≥ 0;

{5x - 4 ≥ 0;

{12x + 1 ≥ 0; {x ≥ - 1/2;

{x ≥ 4/5;

{x ≥ - 1/12. x ∈ [0,8; ∞).

2. Возводим в квадрат:

√ (2x + 1) + √ (5x - 4) = √ (12x + 1); 2x + 1 + 2√ (2x + 1) √ (5x - 4) + 5x - 4 = 12x + 1; 2√ (2x + 1) √ (5x - 4) + 7x - 3 = 12x + 1; 2√ (2x + 1) (5x - 4) = 12x + 1 - 7x + 3; 2√ (10x^2 - 3x - 4) = 5x + 4.

3. Снова в квадрат:

{5x + 4 ≥ 0;

{4 (10x^2 - 3x - 4) = (5x + 4) ^2; {x ≥ - 4/5 ∈ ОДЗ;

{40x^2 - 12x - 16 = 25x^2 + 40x + 16; 15x^2 - 52x - 32 = 0; D/4 = 26^2 + 15 * 32 = 676 + 480 = 1156 = 34^2; x = (26 ± 34) / 15; x = (26 - 34) / 15 = - 8/15 ∉ ОДЗ; x = (26 + 34) / 15 = 60/15 = 4 ∈ ОДЗ.

Ответ: 4.