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:

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

Answer 2

To solve the equation (-y³ + 5y)2y - 10y²(1 + 0.2y²) = -4y, we'll simplify and solve step by step. Let's go through it:

(-y³ + 5y)2y - 10y²(1 + 0.2y²) = -4y

Expanding the equation, we have:

(-y³ + 5y)2y - 10y² - 2y²(0.2y²) = -4y

Next, let's simplify further:

(-2y⁴ + 10y²) - 10y² - 0.4y⁴ = -4y

Combining like terms:

-2y⁴ - 0.4y⁴ + 10y² - 10y² + 4y = 0

Simplifying the equation:

-2.4y⁴ + 4y = 0

Now, let's factor out common terms:

y(-2.4y³ + 4) = 0

To solve this equation, we set each factor equal to zero and solve for y:

1. y = 0

2. -2.4y³ + 4 = 0

Let's solve the second equation:

-2.4y³ + 4 = 0

Adding 2.4y³ to both sides:

2.4y³ = 4

Dividing by 2.4:

y³ = 4/2.4

y³ = 1.6667

Taking the cube root of both sides:

y = ∛(1.6667)

y ≈ 1.123

Therefore, the solutions to the equation are y = 0 and y ≈ 1.123.