Question

VA ROG MULT AJUTORRR
(multimea nr reale)​

Answer 1

Răspuns: a) 4

                b) 2

                c) 0 sau 4

a) 4 - $\sqrt{4}$ = 2

   4 - 2 = 2

   2 = 2 Adevarat

b) $\sqrt{2x}$ + $\sqrt{6 - x}$ = 4

    $\sqrt{4}$ + $\sqrt{6 - 2}$ = 4

     2 + $\sqrt{4}$ = 4

     2 + 2 = 4

     4 = 4 Adevarat

c)

$I. \sqrt{1 + 0} + \sqrt{4 - 0} = 3\\\sqrt{1} + \sqrt{4} = 3\\1 + 2 = 3\\3 = 3 Adevarat\\\\II. \sqrt{1 + 8} + \sqrt{4 - 4 } = 3\\\\\sqrt{9} + \sqrt{0} = 3\\3 + 0 = 3\\3 = 3 Adevarat$

Bafta!

Answer 2

Explicație pas cu pas:

a)

$x \geqslant 0$

$x - \sqrt{x} = 2 \iff x - 2 = \sqrt{x}$

$\sqrt{x} \geqslant 0 \implies x - 2 \geqslant 0 \implies x \geqslant 2 \\ \implies x \in \Big[2 ; +\infty \Big)$

${(x - 2)}^{2} = {( \sqrt{x} )}^{2}$

${x}^{2} - 4x + 4 = x \iff {x}^{2} - 5x + 4 = 0$

$(x - 1)(x - 4) = 0$

$x - 1 = 0 \implies x = 1 \not\in \Big[2 ; +\infty \Big)$

$x - 4 = 0 \implies \bf x = 4 \in \Big[2 ; +\infty \Big)$

6$\implies S = \Big\{4\Big\}$

b)

$\sqrt{2x} + \sqrt{6 - x} = 4$

$x \geqslant 0$

$6 - x \geqslant 0 \implies x \leqslant 6$

$\implies x \in \Big[0 ;6 \Big]$

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

$2x + 2 \sqrt{2x(6 - x)} + 6 - x = 16$

$2 \sqrt{2x(6 - x)} = 10 - x$

${(2 \sqrt{2x(6 - x)})}^{2} = {(10 - x)}^{2}$

$4 \cdot 2x(6 - x) = 100 - 20x + {x}^{2}$

$48x - 8 {x}^{2} = 100 - 20x + {x}^{2}$

$9{x}^{2} - 68x + 100 = 0$

$(9x - 50)(x - 2) = 0$

$9x - 50 = 0 \implies \bf x = \dfrac{50}{9} \in \Big[0 ;6 \Big]$

$x - 2 = 0 \implies \bf x = 2 \in \Big[0 ;6 \Big]$

$\implies S = \Big\{ 2; \dfrac{50}{9} \Big\}$

c)

$\sqrt{1 + 2x} + \sqrt{4 - x} = 3$

$1 + 2x \geqslant 0 \implies x \geqslant \dfrac{1}{2}$

$4 - x \geqslant 0 \implies x \leqslant 4$

$\implies x \in \Big[ - \dfrac{1}{2} ;4 \Big]$

${(\sqrt{1 + 2x} + \sqrt{4 - x})}^{2} = {3}^{2}$

$1 + 2x + 4 - x + 2 \sqrt{(1 + 2x)(4 - x)} = 9$

$2 \sqrt{(1 + 2x)(4 - x)} = 4 - x$

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

$4(4 - x + 8x - 2 {x}^{2}) = 16 - 8x + {x}^{2}$

$16 + 28x - 8 {x}^{2} = 16 - 8x + {x}^{2}$

$9 {x}^{2} - 36x = 0$

$9x(x - 4) = 0$

$x = 0 \in \Big[ - \dfrac{1}{2} ;4 \Big]$

$x - 4 = 0 \implies x = 4 \in \Big[ - \dfrac{1}{2} ;4 \Big]$

$\implies S = \Big\{ 0; 4 \Big\}$