Question

(x^2+5x) / x-3 = (x-27) / 3-x

Answer 1

(х²+5х)/(х-3)=(х-27)/(3-х)

(х²+5х)/(х-3)=-(х-27)/(х-3)|×(х-3)

при условии, что (х-3)≠0 так как на "0" делить нельзя.
х-3≠0
х≠3.

(х²+5х)=-(х-27)
(х²+5х)+(х-27)=0
х²+5х+х-27=0
х²+6х-27=0
По теореме Виета:
х1+х2=-6
х1×х2=-27
х1=-9
х2=3-не является решением нашего уравнения , по условию ≠3.

х=-9.

Проверка:

(9²+5×(-9))/((-9)-3)=((-9)-27)/(3-(-9))
(81-45)/(-12)=(-46)/12
46/(-12)=(-46)/12
-3(10/12)=-3(10/12)
-3(5/6)=-3(5/6)-истина.

Ответ: х=-9

Answer 2

To solve the given equation, let's first simplify both sides of the equation and then find the value of 'x' that satisfies the equation.

Given equation:

(x^2 + 5x) / (x - 3) = (x - 27) / (3 - x)

Step 1: Simplify the fractions.

(x^2 + 5x) / (x - 3) can be simplified using factoring:

x(x + 5) / (x - 3)

(x - 27) / (3 - x) can be simplified by changing the sign of the denominator:

-(x - 27) / (x - 3)

So, the equation becomes:

x(x + 5) / (x - 3) = -(x - 27) / (x - 3)

Step 2: Eliminate the denominator by multiplying both sides by (x - 3):

x(x + 5) = -(x - 27)

Step 3: Expand and simplify the equation:

x^2 + 5x = -x + 27

Step 4: Bring all the terms to one side of the equation to set it to zero:

x^2 + 6x - 27 = 0

Step 5: Factor the quadratic equation (if possible) or use the quadratic formula to find the values of 'x':

This quadratic equation doesn't factor easily, so let's use the quadratic formula:

The quadratic formula is given by:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 1, b = 6, and c = -27.

x = (-(6) ± √((6)^2 - 4 * 1 * (-27))) / 2 * 1 x = (-6 ± √(36 + 108)) / 2 x = (-6 ± √144) / 2 x = (-6 ± 12) / 2

Now, solve for two values of 'x':

1. x = (-6 + 12) / 2 x = 6 / 2 x = 3

2. x = (-6 - 12) / 2 x = -18 / 2 x = -9

So, the solutions to the equation are x = 3 and x = -9.