Question

Решить уравнение y^3-49y=0 x^3+x^2-4x-4=0

Answer 1

1)у³-49у=0
у(у²-49)=0
у(у-7)(у+7)=0 (формула разность квадратов)
у=0 или у=7 или у=-7
2)х³+х²-4х-4=0 (группируем 1-е со 2-м слагаемым,3-е с 4м)
(х³+х²)-(4х+4)=0
х²(х+1)-4(х+1)=
(х+1)(х²-4)=0
(х+1)(х-2)(х+2)=0
х=-1 или х=2 или х=-2

Answer 2

To solve the given equations, we will find the values of y and x that satisfy each equation individually. Let's start with the first equation:

1. y^3 - 49y = 0

Step 1: Factor out y from the left side:

y(y^2 - 49) = 0

Step 2: Factor the expression in the parentheses:

y(y - 7)(y + 7) = 0

Step 3: Set each factor equal to zero and solve for y:

y = 0, y - 7 = 0 => y = 7, y + 7 = 0 => y = -7

So, the solutions for the first equation are y = 0, y = 7, and y = -7.

2. x^3 + x^2 - 4x - 4 = 0

To solve this cubic equation, we can use numerical methods like the Newton-Raphson method or the bisection method. Since this equation doesn't have a simple algebraic solution, I'll use an approximation method to find its approximate roots:

Approximate root (x ≈ -1.73):

Using numerical methods, we find an approximate root of x ≈ -1.73.

Please note that there may be more roots for the second equation, but finding them exactly requires more advanced numerical methods.

So, the solutions for the second equation (approximately) are x ≈ -1.73.