Question

X^3-(a+b+c)x^2+(ab+bc+ac)x-abc=0​

Answer 1

Ответ:

x = a ; x=b ; x = c

Объяснение:

Уравнение 3 степени имеет не более 3 корней , проверкой

убеждаемся , что  a , b , c - корни данного уравнения и других

быть не может .

Проверка для x = a  , для остальных все аналогично :

a³-(a+b+c)·a² + (ab+bc+ac)·a -abc = a³- a³-a²b-a²c + a²b +abc +a²c -

abc = 0 (верно)

вариант 2 :

легко убедиться , что левая часть уравнения равна

(x-a)(x-b)(x-c)  (   ну просто раскройте скобки)

(x-a)(x-b)(x-c) = 0 ⇔ x =a  или x=b  или x =c

вариант 3 :  a ; b ; c  - корни по теореме , обратной к теореме

Виета для уравнения 3 степени

вариант 4 :

x³ - x²(b+c) +xbc -ax² + (b+c)ax -abc =0

x(x² -(b+c)x +bc) -a(x² - (b+c)x +bc) =0

(x - a)(x² -(b+c)x +bc) = 0

x = a  или  x² -(b+c)x +bc = 0

x² -(b+c)x +bc = 0  ⇔ x =b или  x = c (  по теореме , обратной к теореме Виета )

Answer 2

This is a cubic equation of the form:

x^3 - (a + b + c)x^2 + (ab + bc + ac)x - abc = 0

To solve for x, we can use different methods such as:

1. Factorization: If we can factor the left-hand side of the equation, we can find the roots. However, this cubic equation is not always factorable.

2. Cardano's formula: This is a formula for finding the roots of a cubic equation. However, it can be quite complicated to use.

3. Numerical methods: We can use numerical methods such as the Newton-Raphson method or the bisection method to find the roots of the equation.

Here, we will use a simpler method called the Rational Root Theorem to find one of the possible rational roots of the equation.

The Rational Root Theorem states that if a polynomial has integer coefficients, then any rational root of the polynomial must be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

For this equation, the constant term is -abc and the leading coefficient is 1. Therefore, any rational root must be of the form p/q, where p is a factor of abc and q is a factor of 1.

Possible factors of abc are: ±1, ±a, ±b, ±c, ±ab, ±bc, ±ac, and ±abc.

Possible factors of 1 are: ±1.

Therefore, the possible rational roots are:

±1, ±a, ±b, ±c, ±ab, ±bc, ±ac, and ±abc.

We can test each of these possible roots by substituting them into the equation and seeing if the left-hand side equals zero.

Once we find one root, we can use polynomial division or synthetic division to reduce the equation to a quadratic equation and solve for the remaining roots.

Note that this method only gives us the rational roots of the equation. There may be irrational or complex roots that we cannot find using this method.