Question

3) (-y³ + 5y)2y - 10y²(1 +0,2y²)=-4y¹;​

Answer 1

Ответ:

$y_1=0,\\y_2=\sqrt[3]{-\dfrac{4}{5}}\approx-0.92831776672\ldots$

Объяснение:

$(-y^3+5y)2y-10y^2(1+0.2y^2)=-4y\\-3y^4+10y^2-10y^2-2y^4=-4y\\-5y^4-4y=0\\-y(5y^3+4)=0\implies\left[\begin{aligned}-y&=0\\5y^3+4&=0\end{aligned}\right.\implies\left[\begin{aligned}y&=0\\5y^3&=-4\implies y^3=-\dfrac{4}{5}\implies y=\sqrt[3]{-\dfrac{4}{5}}\end{aligned}\right.\\y_1=0,\\y_2=\sqrt[3]{-\dfrac{4}{5}}\approx-0.92831776672\ldots$