Предмет: Алгебра, автор: jungsoojung

решите уравнение 5 корень 128у^2 + 5 корень 64у = 24

Ответы

Автор ответа: Аноним

0

$5 sqrt{128y^2} +5 sqrt{64y} =24$
ОДЗ: $y geq 0$
Воспользуемся свойством степеней
$5 sqrt{(8 sqrt{2}y)^2 } +5 sqrt{64y} =24 \ \ 5|8 sqrt{2} y|+5 sqrt{64y} =24 \ 40y sqrt{2} +5 sqrt{64y} =24 \ \ 5 sqrt{64y} =24-40y sqrt{2}$
Возведем оба части до квадрата
$25cdot64y=(24-40y sqrt{2} )^2 \ 1600y=3200y^2-1920y sqrt{2} +576 \ 3200y^2-(1660+1920 sqrt{2} )y+576=0$
Находим дискриминант
$D=b^2-4ac=(-1600-1920 sqrt{2} )^2-4cdot3200cdot576 \ D=25600000+6144000 sqrt{2}$
Берем корень положительный, потому что отрицательный корень не будет удовлетворять ОДЗ
$y= dfrac{1600+1920 sqrt{2}- sqrt{2560000+6144000 sqrt{2} } }{6400}$

Автор ответа: jungsoojung

0

спасибо огромное ;)

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