Question

Giải phương trình:

a) \(\sqrt{x^2-4x+4}+\left|x-4\right|=0\)

b) \(\sqrt{x^2-1}+1=x^2\)

c) \(\sqrt{x+5}=1-x\)

d) \(\sqrt{x^2-2x+1}=\sqrt{6+4\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)

Answer 1

d) \(\sqrt{x^2-2x+1}\) = \(\sqrt{6+4\sqrt{2}}\) - \(\sqrt{6-4\sqrt{2}}\)

⇔ \(\sqrt{x^2-2x+1}\) = \(\sqrt{\left(2+\sqrt{2}\right)^2}\) - \(\sqrt{\left(2-\sqrt{2}\right)^2}\)

⇔ \(\sqrt{x^2-2x+1}\) = 2 + \(\sqrt{2}\) - 2 +\(\sqrt{2}\)

⇔ \(\sqrt{x^2-2x+1}\) = \(2\sqrt{2}\)

⇔ x² - 2x +1 = 8

⇔ x² - 2x +1 -8 = 0

⇔ x² - 2x - 7 = 0

x₁ = 1 + \(2\sqrt{2}\) (nhận)

x₂ = 1 - \(2\sqrt{2}\) (nhận)

Vậy S = \(\left\{1+2\sqrt{2};1-2\sqrt{2}\right\}\)

Answer 2

c) \(\sqrt{x+5}\) = 1 - x

⇔ x + 5 = 1 - 2x + x²

⇔ 1 - 2x + x² - x - 5 = 0

⇔ x² - 3x - 4 = 0

x₁ = 4 (nhận)

x₂ = -1 (nhận)

Vậy S = {4;-1}