Question

X+5/x-2+x-5/x+2=10x+8/x^2-4

Answer 1

Ответ:

Объяснение: Общий множитель (x-2)(x+2);

(x-5)(x+2)+(x-5)(x-2)=10x+8;

x^2+2x-5x-10+x^2-2x-5x+10=10x+8(убираемпротивоположные значения);

x^2-5x+x^2-5x=10x+8;

2x^2-10x=10x+8;

2x^2-10x-10x=8

2x^2-20x=8 (поделим обе части уравнения на 2)

x^2-10x=4

X^2-10x-4=0

D=(-10)^2-4(-4)= 84

x1=5-√29

x2=5+√29

Answer 2

To solve the given equation:

x + 5/(x - 2) + x - 5/(x + 2) = (10x + 8)/(x^2 - 4)

Let's simplify the equation step by step.

First, we'll work on the denominators (x - 2) and (x + 2) in the fractions:

(x(x - 2) + 5)/(x - 2) + (x(x + 2) - 5)/(x + 2) = (10x + 8)/(x^2 - 4)

(x^2 - 2x + 5)/(x - 2) + (x^2 + 2x - 5)/(x + 2) = (10x + 8)/(x^2 - 4)

Now, let's find a common denominator for the fractions on the left side. The common denominator is (x - 2)(x + 2) = x^2 - 4.

[(x^2 - 2x + 5)(x + 2) + (x^2 + 2x - 5)(x - 2)]/(x^2 - 4) = (10x + 8)/(x^2 - 4)

Simplifying the numerators:

[(x^3 + 2x^2 - 4x + 2x^2 - 4x + 10) + (x^3 - 2x^2 - 5x + 2x^2 - 4x + 10)]/(x^2 - 4) = (10x + 8)/(x^2 - 4)

Combining like terms:

(2x^3 - 6x + 20)/(x^2 - 4) = (10x + 8)/(x^2 - 4)

Since both sides of the equation have a common denominator of (x^2 - 4), we can cancel it out:

2x^3 - 6x + 20 = 10x + 8

Rearranging the equation:

2x^3 - 16x + 12 = 0

Now we have a cubic equation. To solve it, we can use numerical methods or factoring techniques. Unfortunately, the equation doesn't seem to have any obvious factorizations, so we'll resort to numerical methods or calculators to find the approximate solutions.