Question

(2y³ + 3y² - 7) - (5 + 3y + y³) = 3y² + y³ - 5yрешите уравнение

Answer 1

Избавимся от лишних скобок
${2y}^{3} + {3y}^{2} - 7 - (5 + 3y + {y}^{3} ) = 3y ^{2} + {y}^{3} - 5y$
если перед скобкой стоит знак минус , то в скобках знаки меняются на противоположные
${2y}^{3} + {3y}^{2} - 7 - 5 - 3y - {y}^{3} = {3y}^{2} + {y}^{3} - 5y$
убрать равные слогаемые
${2y}^{3} - 7 - 5 - 3y - y^{3} = {y}^{ 3} - 5y$
${y}^{3} - 12 - 3y = {y}^{3} - 5y$
$- 12 - 3y = 5y$
$- 3y + 5y = 12$
$2y = 12$
$y = 6$

Answer 2

Let's begin by simplifying the left-hand side of the equation:

(2y³ + 3y² - 7) - (5 + 3y + y³) = 2y³ + 3y² - 7 - 5 - 3y - y³ (distributing the negative sign) = y³ + 2y³ + 3y² - 3y - 12 (combining like terms)

Now we can substitute this expression into the original equation and solve for y:

(y³ + 2y³ + 3y² - 3y - 12) = 3y² + y³ - 5y

Simplifying:

3y³ + 3y² - 5y - 12 = 3y² + y³ - 5y

The y² and y terms cancel out, leaving:

2y³ - 12 = 0

Dividing both sides by 2:

y³ - 6 = 0

Adding 6 to both sides:

y³ = 6

Taking the cube root of both sides:

y = ∛6

Therefore, the solution to the equation is y = ∛6.