Question

rezolvați ecuațile rpd dau coroană​

Answer 1

a)

${x}^{2} = 1 \iff |x| = 1 \implies x = \pm 1$

$\implies S = \{-1;1\}$

b)

$2 {x}^{2} = 12 \iff {x}^{2} = 6$

$|x| = \sqrt{6} \implies x = \pm \sqrt{6}$

$\implies S = \{ - \sqrt{6} ; \sqrt{6} \}$

c)

${(x - 2)}^{2} = 16 \iff {(x - 2)}^{2} = {4}^{2}$

$|x - 2| = 4$

$x - 2 = - 4 \iff x = - 4 + 2 \implies x = - 2$

$x - 2 = 4 \iff x = 4 + 2 \implies x = 6$

$\implies S = \{-2;6\}$

d)

$2 {x}^{2} - 2 = 6 \iff 2 {x}^{2} = 8$

${x}^{2} = 4 \iff {x}^{2} = {2}^{2} \iff x = \pm 2$

$\implies S = \{-2;2\}$

e)

${(2x + 1)}^{2} = 25 \iff {(2x + 1)}^{2} = {5}^{2}$

$|2x + 1| = 5$

$2x + 1 = - 5 \iff 2x = - 5 - 1 \\ 2x = - 6 \implies x = - 3$

$2x + 1 = 5 \iff 2x = 5 - 1 \\ 2x = 4 \implies x = 2$

$\implies S = \{-3;3\}$

f)

${(2x - 3)}^{2} = 36 \iff {(2x - 3)}^{2} = {6}^{2}$

$|2x - 3| = 6$

$2x - 3 = - 6 \iff 2x = - 6 + 3 \\ 2x = - 3 \iff x = - \dfrac{3}{2}$

$2x - 3 = 6 \iff 2x = 6 + 3 \\ 2x = 9 \iff x = \dfrac{9}{2}$

$\implies S = \Big\{- \dfrac{3}{2} ; \dfrac{9}{2} \Big\}$