1)5y^2-4y=0
2)2x^2+3x=0
3)1-4y^2=0

Предыдущий
вопрос Следующий
вопрос

1) 5y^2 - 4y = 0
=> y(5y - 4) = 0

This equation can be separated into two equations:
y = 0
5y - 4 = 0
5y = 4
y = 4/5

Therefore, the solutions are y = 0, y = 4/5.

2) 2x^2 + 3x = 0
=> x(2x + 3) = 0

This equation can be separated into two equations:
x = 0
2x + 3 = 0
2x = -3
x = -3/2

Therefore, the solutions are x = 0, x = -3/2.

3) 1 - 4y^2 = 0
=> 4y^2 = 1
=> y^2 = 1/4
=> y = ±1/2

Therefore, the solutions are y = 1/2, y = -1/2.