Question

решите уравнение
√17+2х-3х^2=х+1​

Answer 1

Ответ:

$\sqrt{17 + 2x - {3x}^{2} } = x + 1$

${ \sqrt{17 + 2x - {3x}^{2} } }^{2} = {(x + 1)}^{2}$

$17 + 2x - {3x}^{2} = {x}^{2} + 2x \times 1 + {1}^{2}$

$17 + 2x - {3x}^{2} = {x}^{2} + 2x + 1$

$17 - {3x}^{2} = {x}^{2} + 1$

$- {3x}^{2} - {x}^{2} = 1 - 17$

$- 3 {x}^{2} - 1 {x}^{2} = 1 - 17$

$- 4 {x}^{2} = 1 - 17$

$- 4 {x}^{2} = - 16$

$- 4 {x}^{2} = - 16| \div - 4$

${x}^{2} = 4$

$x = ± \sqrt{4}$

$x = ± \sqrt{ {2}^{2} }$

$x = ±2$

$x = - 2 \\ x = 2$

$\sqrt{17 + 2 \times ( - 2) - 3 \times {( - 2)}^{2} } = - 2 + 1 \\ \sqrt{17 + 2 \times 2 - 3 \times {2}^{2} } = 2 + 1$

$\sqrt{17 + 2 \times ( - 2) - 3 \times {( - 2)}^{2} } = - 2 + 1 \\ \sqrt{17 - 4 - 3 \times {2}^{2} } = - 1 \\ \sqrt{17 - 4 - 3 \times 4} = - 1 \\ \sqrt{17 - 4 - 12} = - 1 \\ \sqrt{1} = - 1 \\ 1 = - 1$

$\sqrt{17 + 2 \times 2 - 3 \times {2}^{2} } = 2 + 1 \\ \sqrt{17 + 4 - 3 \times 4} = 3 \\ \sqrt{17 + 4 - 12} = 3 \\ \sqrt{9} = 3 \\ \sqrt{ {3}^{2} } = 3 \\ 3 = 3$

$1 = - 1 \\ 3 = 3$

$x≠ - 2 \\ x = 2$

$x = 2$