Question

Calculați...................

Answer 1

$12) {3}^{x} \times {5}^{x} = 15$

${3}^{x} \times {5}^{x} = {3}^{1} \times {5}^{1} = > x = 1$

$13) \sqrt{x - 1} = \sqrt{ {x}^{2} - x - 2} \: | {( \: )}^{2}$

$x - 1 \geqslant 0$

${x}^{2} - x - 2 \geqslant 0$

$x - 1 = {x}^{2} - x - 2$

$- {x}^{2} + x + x - 1 + 2 = 0$

$- {x}^{2} + 2x + 1 = 0 \: | \times ( - 1)$

${x}^{2} - 2x - 1 = 0$

$a = 1$

$b = - 2$

$c = - 1$

$\Delta = {b}^{2} - 4ac$

$\Delta = {( - 2)}^{2} - 4 \times 1 \times ( - 1)$

$\Delta = 4 + 4$

$\Delta = 8$

$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a} = \frac{ - ( - 2) \pm \sqrt{8} }{2 \times 1} = \frac{2 \pm2 \sqrt{2} }{2}$

$x_{1}= \frac{2 + 2 \sqrt{2} }{2} = \frac{2(1 + \sqrt{2}) }{2} = 1 + \sqrt{2}$

$x_{2}= \frac{2 - 2 \sqrt{2} }{2} = \frac{2(1 - \sqrt{2} )}{2} = 1 - \sqrt{2}$

$14) \frac{ {2}^{x} }{ {3}^{x} } = \frac{3}{2}$

$\frac{ {2}^{ - 1} }{ {3}^{ - 1} } = \frac{ \frac{1}{2} }{ \frac{1}{3} } = \frac{1}{2} \times 3 = \frac{3}{2} = > x = - 1$

$15) log_{5}(3x + 1) = 1 + log_{5}(x - 1)$

$log_{5}(3x + 1) = log_{5}(5) + log_{5}(x - 1)$

$log_{5}(3x + 1) = log_{5}(5(x - 1))$

$3x + 1 = 5(x - 1)$

$3x + 1 = 5x - 5$

$3x - 5x = - 5 - 1$

$- 2x = - 6 \: | \div ( - 2)$

$x = 3$

$16) \frac{1}{ {2}^{x} } = \frac{ {4}^{x} }{8}$

${2}^{x} \times {4}^{x} = 8 \times 1$

${2}^{x} \times {2}^{2x} = {2}^{3}$

${2}^{x + 2x} = {2}^{3}$

${2}^{3x} = {2}^{3} = > x = 1$

$17) \sqrt{2x + 3} = x + 2 \: | {( \: )}^{2}$

$2x + 3 \geqslant 0$

$2x + 3 = {x}^{2} + 4x + 4$

$- {x}^{2} + 2x - 4x + 3 - 4 = 0$

$- {x}^{2} - 2x - 1 = 0\: | \times ( - 1)$

${x}^{2} + 2x + 1 = 0$

${(x + 1)}^{2} = 0$

$x + 1 = 0 = > x = - 1$

$18) \sqrt{ {x}^{2} + 2x - 3} = 2 \sqrt{3} \: | {( \: )}^{2}$

${x}^{2} + 2x - 3\geqslant0$

${x}^{2} + 2x - 3 = 12$

${x}^{2} + 2x - 3 - 12 = 0$

${x}^{2} + 2x - 15 = 0$

${x}^{2} + 5x - 3x - 15 = 0$

$x(x + 5) - 3(x + 5) = 0$

$(x - 3)(x + 5) = 0$

$x - 3 = 0 = > x = 3$

$x + 5 = 0 = > x = - 5$

$19) log_{2}( \sqrt{x + 1} ) = 1$

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

$\sqrt{x + 1} = 2 \: | {( \: )}^{2}$

$x + 1 \geqslant 0$

$x + 1 = 4$

$x = 4 - 1$

$x = 3$

$20) \sqrt[3]{ {x}^{3} + x + 1 } = x \: | {( \: )}^{3}$

${x}^{3} + x + 1 = {x}^{3}$

$x + 1 = 0$

$x = - 1$