Решите иррациональное уравнение :
√(x²-3x-10)+√(x²+3x+2)=√(x²+8x+12)
С ОДЗ!!!
Ответы
√(x²-3x-10)+√(x²+3x+2)=√(x²+8x+12)
найдем одз
D = 9 + 40 = 49 x12 = (3 +- 7)/2 = 5 -1 (x + 2)(x - 5) >= 0
x ∈ (-∞, -2] U [5, +∞)
D = 9 - 8 = 1 x12 = (-3 +- 1)/2 = -1 -2 (x + 1)(x + 2) >= 0
x ∈ (-∞, -2] U [-1, +∞)
D = 84 - 48 = 16 x12 = (-8 +- 4)/2 = - 6 -2 (x + 2)(x + 6) >= 0
x ∈ (-∞, -6] U [-2, +∞)
общее x∈(-∞,-6] U {-2} U [5,+∞)
√(x²-3x-10)+√(x²+3x+2)=√(x²+8x+12)
√ (x + 2)(x - 5) + √(x + 1)(x + 2) = √(x + 2)(x + 6)
x = -2 корень "убираем" его
√(x - 5) + √(x + 1) = √(x + 6)
x - 5 + 2√(x - 5)√(x + 1) + x + 1 = x + 6
2√ (x - 5)√(x + 1) = 10 - x (x <= 10 x>=5)
4(x² - 4x - 5) = 100 - 20x + x²
4x² - 16x - 20 = 100 - 20x + x²
3x² + 4x - 120 = 0
D = 16 + 4*3*120 = 1456 = 16*91
x12 = (-4 +- 4√91)/6 = (-2 +- 2√91)/3 ≈ 5.7 -7
x1 = (-2 - 2√91)/3 <= 5 нет
x2 = (-2 + 2√91)/3 >=5 и <= 10 корень
ответ х = { (-2 + 2√91)/3, -2}