Question

Mă puteți ajuta, vă rog, cu acest exercițiu ?

Answer 1

a)

$\sqrt{x^2-1} = 2\Rightarrow x^2-1 = 4\\ x^2 = 5\\ x = \pm \sqrt{5}$

b)

$\sqrt{3-2x^2}=1\Rightarrow 3 - 2x^2 = 1\\2 = 2x^2\\ x^2 = 1\\ x = \pm\sqrt{1} = \pm 1$

c)

$\sqrt{6x-4} = 0\Rightarrow 6x-4 = 0 \\ 6x = 4\\ x = \frac{4}{6} = \frac{2}{3}$

d)

$\sqrt{2x-3} = 5\Rightarrow 2x-3 = 25 \\ 2x = 28\\ x = 14$

e)

$\sqrt{7x-2} = -3 \Rightarrow S = \varnothing$

f)

$\sqrt{|x-3|+2} = 3 \Rightarrow |x-3|+2 = 9\\ |x-3| = 7\\ x-3 = \pm 7 \Rightarrow S = \{-4, 10\}$

g)

$\sqrt{5 - |1 - x^2|} = 2 \Rightarrow 5 - |1 - x^2| = 4\\1 = |1 - x^2|\Rightarrow\\(1) x^2 = 0 \Rightarrow x = \pm \sqrt{0} = \pm 0 = 0\\(2) 1-x^2 = -1 \Rightarrow x^2 = 2\Rightarrow x = \pm \sqrt{2}\\S = \{-\sqrt{2}, 0, \sqrt{2}\}$

h)

$\sqrt{5 + \sqrt{x-1}} = 3\Rightarrow 5 + \sqrt{x-1} = 9 \Rightarrow \sqrt{x-1}= 4\Rightarrow x-1 = 16 \Rightarrow x = 17$

i)

$\sqrt{7 - \sqrt{x+1}} = 2 \Rightarrow 7 - \sqrt{x+1} = 4\Rightarrow \sqrt{x+1} = 3\Rightarrow x+1 = 9\Rightarrow x = 8$