Question

Допоможіть будь ласка ​

Answer 1

Ответ:

5.

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

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

$1 - x = 4 {x}^{2} + 4x + 1$

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

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

$- 1x - 4 {x}^{2} - 4x = 0$

$- 5x - {4x}^{2} = 0$

$- x(5 + 4x) = 0$

$x(5 + 4x) = 0$

$x = 0 \\ 5 + 4x = 0$

$5 + 4x = 0 \\ 4x = - 5 \\ \div 4 \\ x = - \frac{5}{4}$

$x = 0 \\ x = - \frac{5}{4}$

$\sqrt{1 - 0} = 2 \times 0 + 1 \\ \sqrt{1 - ( - \frac{5}{4} )} = 2 \times ( - \frac{5}{4} ) + 1$

$\sqrt{1 - 0} = 2 \times 0 + 1 \\ \sqrt{1} = 2 \times 0 + 1 \\ 1 = 0 + 1 \\ 1 = 1$

$\sqrt{1 - ( - \frac{5}{4} )} = 2 \times ( - \frac{5}{4} ) + 1 \\ \sqrt{1 + \frac{5}{4} } = - \frac{5}{2} + 1 \\ \sqrt{ \frac{9}{4} } = - \frac{3}{2} \\ \frac{3}{2} = - \frac{3}{2}$

$1 = 1 \\ \frac{3}{2} = - \frac{3}{2}$

$x = 0 \\ x≠ - \frac{5}{4}$

$x = 0$

ответ:Б.один

6.

$lg25 + lg4$

$lg25 \times 4$

$lg100$

$lg {10}^{2}$

$2lg10$

$2 \times 1$

$2$

ответ:А.2