Question

1) x^3-81x=0 2) x^4 - 19x^2+48 =0​

Answer 1

Объяснение:

Это просто. Объяснять не надо

Answer 2

Let's solve each of the equations separately:

1. x^3 - 81x = 0

First, you can factor out an x from both terms:

x(x^2 - 81) = 0

Now, you have a product of two factors that can equal zero. Set each factor equal to zero and solve for x:

1. x = 0
2. x^2 - 81 = 0

For the second equation:

2. x^4 - 19x^2 + 48 = 0

This is a quadratic equation in terms of x^2. You can make a substitution to simplify it:

Let y = x^2. Now, the equation becomes:

y^2 - 19y + 48 = 0

You can then factor this quadratic equation:

(y - 16)(y - 3) = 0

Now, set each factor equal to zero:

1. y - 16 = 0 y = 16

2. y - 3 = 0 y = 3

Now, recall the substitution y = x^2 and solve for x:

1. x^2 = 16 Take the square root of both sides: x = ±4

2. x^2 = 3 Take the square root of both sides: x = ±√3

So, for equation 2, you have four solutions: x = 4, x = -4, x = √3, and x = -√3.