$\sqrt{2x+15} - \sqrt{2x-1} = \frac{10}{ \sqrt{2x-1} }$

Question

Математика

Saganvanb

6 год назад

$\sqrt{2x+15} - \sqrt{2x-1} = \frac{10}{ \sqrt{2x-1} }$

ОТВЕТЫ

Neosab

Jul 7, 2019

$amp;#10; \sqrt{2x+15} - \sqrt{2x-1} = \frac{10}{ \sqrt{2x-1} }$
ОДЗ:
$\left \{ {{2x+15 \geq 0} \atop {2x-1\ \textgreater \ 0}} \right.$
$\left \{ {{2x \geq -15} \atop {2x\ \textgreater \ 1}} \right.$
$\left \{ {{x \geq -7.5} \atop {x\ \textgreater \ 0.5}} \right.$
$x$ ∈ $(0.5;+$ ∞ $)$

$\sqrt{2x+15}* \sqrt{2x-1} - (\sqrt{2x-1})^2 = 10$
$\sqrt{(2x+15)(2x-1)}- (2x-1}) = 10$
$\sqrt{4x^2+28x-15} = 10+2x-1$
$\sqrt{4x^2+28x-15} =2x+9$
$\left \{ {{2x+9 \geq 0} \atop {(\sqrt{4x^2+28x-15})^2 =(2x+9)^2}} \right.$
$\left \{ {{2x \geq 9} \atop {4x^2+28x-15} =4x^2+36x+81} \right.$
$\left \{ {{x \geq -4.5} \atop {8x=-96} \right.$
$\left \{ {{x \geq -4.5} \atop {x=-12} \right.$
Ответ: нет корней

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