Question

ПОМОГИТЕ ПОЖАЛУЙСТА СРОЧНО!

Answer 1

Ответ:

$\frac{2}{ \sqrt{2 - x} } = \sqrt{ \frac{x + 6}{x + 3} }$

${(2 - x)}^{ \frac{1}{2} } ( \frac{2}{ \sqrt{2 - x} } ) = {(2 - x)}^{ \frac{1}{2} } \sqrt{ \frac{x + 6}{x + 3} }$

$2 = {( \frac{x + 6}{x + 3} )}^{ \frac{1}{2} } \times {(2 - x)}^{ \frac{1}{2} }$

$4 = ( \frac{x}{x + 3} + \frac{6}{x + 3} )(2 - x)$

$(x + 3) \times 4 = (x + 3) \times (( \frac{x}{x + 3} + \frac{6}{x + 3} ) \times (2 - x))$

$4x + 12 = (x + 3) \times (( \frac{x}{x + 3} + \frac{6}{x + 3} ) \times (2 - x))$

$(x + 3)(4x + 12) = (x + 3) \times ( - \frac{ {7x}^{2} }{x + 3} ) + (x + 3) \times ( - \frac{ {x}^{3} }{x + 3} ) + (x + 3) \times ( \frac{36}{x + 3} )$

$(4x + 12)(x + 3) = ( - 1) {x}^{3} - 7 {x}^{2} + 36$

$4 {x}^{2} + 24x + 36 + ( {x}^{3} + 7 {x}^{2} - 36) = - {x}^{3} - 7 {x}^{2} + 36 + ( {x}^{3} + 7 {x}^{2} - 36)$

${x}^{3} + 11 {x}^{2} + 24x = 0$

${x}^{3} + 11 {x}^{2} + 24x = 0$

$x \times ( {x}^{2} + 11x + 24)$

${x}^{2} + 11x + 24 = 0$

$( {x}^{2} + 11x + 24)$

$[A=1, B=11, C=24]$

$|AC|=|1\times{24}|=24$

${1,2,3,4,6,8,12,24}$

${1\times{24,2}\times{12,3}\times{8,4}\times{6}}$

$3 + 8 = 11$

${x}^{2} + (3x + 8x) + 24$

$(x(x + 3)) + (8x + 24)$

$x(x + 3) + (8(x + 3))$

$x(x + 3) + 8(x + 3)$

$(x + 8)(x + 3)$

$x(x + 8)(x + 3)$

$x = 0$

$x + 3 = 0$

$x + 3 - 3 = 0 - 3$

$x = - 3$

$x + 8 = 0$

$x + 8 - 8 = 0 - 8$

$x = - 8$

$\sqrt{2 - 1x} ≠0$

$\sqrt{2 + ( - 1)( - 8)}$

$\sqrt{10} ≈3.16$

$3.16≠0$

$x + 3≠0$

$- 8 + 3 = - 5$

$- 5≠0$

$0 \leqslant 2 - 1x$

$2 + ( - 1)( - 8) = 10$

$0 \leqslant 10$

$0 \leqslant \frac{x + 6}{x + 3}$

$\frac{ - 8 + 6}{ - 8 + 3}$

$\frac{ - 2}{ - 5}$

$0 \leqslant \frac{2}{5}$

$[]$

$\frac{2}{ \sqrt{2 + 8} } = \sqrt{ \frac{ - 8 + 6}{ - 8 + 3} }$

$\frac{ \sqrt{10} }{5} = \frac{ \sqrt{10} }{5}$

$[]$

$\sqrt{2 - 1x} ≠0$

$\sqrt{2 + ( - 1)( - 3)}$

$\sqrt{5} ≈2.23$

$2.23≠0$

$x + 3≠0$

$- 3 + 3 = 0$

$0 = 0$

$[]$

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$x = - 3$

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-постороннее решение.

$\sqrt{2 - 1x} ≠0$

$\sqrt{2 + ( - 1) \times 0}$

$\sqrt{2 + 0}$

$\sqrt{2} ≈1.41$

$1.41≠0$

$x + 3≠0$

$0 + 3 = 3$

$3≠0$

$0 \leqslant 2 - 1x$

$2 + ( - 1) \times 0$

$2 + 0 = 2$

$0 \leqslant 2$

$0 \leqslant \frac{x + 6}{x + 3}$

$\frac{0 + 6}{0 + 3}$

$\frac{6}{3}$

Сокращаем на общий делитель 6:

$\frac{1}{2}$

$2$

$0 \leqslant 2$

$\frac{2}{ \sqrt{2 + 0} } = \sqrt{ \frac{0 + 6}{0 + 3} }$

$\sqrt{2} = \sqrt{2}$

$[]$

$x = ( - 8.0)$

$2 = \frac{ \sqrt{2} \sqrt{5} \sqrt{10} }{5}$

$2 = \frac{ \sqrt{2 \times 5 \times 10} }{5}$

$2 = \frac{ \sqrt{100} }{5}$

$2 = \frac{ \sqrt{ {10}^{2} } }{5}$

$2 = \frac{10}{5}$

Сокращаем на общий делитель 5:

$2 = 2$

$x = ( - 8.0)$