Question

Вот второй

Решите уравнения:

Answer 1

$\displaystyle \sqrt{3x-11}=1+\sqrt{x}\\3x-11=1+2\sqrt{x} +x\\-2\sqrt{x} =1+x-3x+11\\-2\sqrt{x} =12-2x\\\sqrt{x} =-6+x\\x=x^2-12x+36\\x-x^2+12x-36=0\\13x-x^2-36=0\\-x^2+13x-36=0\\x^2-13x+36=0\\x^2-4x-9x+36=0\\x(x-4)-9(x-4)=0\\(x-4)(x-9)=0\\x-4=0,x-9=0\\x=4,x=9\\\sqrt{3*4-11}=1+\sqrt{4},\sqrt{3*9-11}=1+\sqrt{9}\\x\neq 4,x=9\\x=9$

Answer 2

√(3x - 11) = 1 + √x

одз x ≥ 0

3x - 11 ≥ 0   x≥11/3

x∈ [11/3,+∞)

(3x - 11) = 1 + 2√x + x

2x - 12 = 2√x

x - 6 = √x

x ≥ 6

x² - 12x + 36 = x

x² - 13x + 36 = 0

D = 13² - 4*36 = 25

x12 = (13 +- 5)/2 = 4   9

x1 = 4  < 6 нет

x2 = 9

√(27 - 11) = 4 = 1 + √9