Question

X+5/x-2+x+2/x+1=21/(x-2)(x+1)

Answer 1

Ответ:

Вот ответ ,надеюсь правильно

Answer 2

To solve the given equation:

$\frac{x+5}{x-2} + \frac{x+2}{x+1} = \frac{21}{(x-2)(x+1)}$

We'll start by finding a common denominator for the fractions on the left side of the equation, which is $(x-2)(x+1)$.

$\frac{(x+5)(x+1)}{(x-2)(x+1)} + \frac{(x+2)(x-2)}{(x-2)(x+1)} = \frac{21}{(x-2)(x+1)}$

Now, we can combine the fractions over the common denominator:

$\frac{(x+5)(x+1) + (x+2)(x-2)}{(x-2)(x+1)} = \frac{21}{(x-2)(x+1)}$

Expanding the numerator:

$\frac{x^2 + 6x + 5 + (x^2 - 4)}{(x-2)(x+1)} = \frac{21}{(x-2)(x+1)}$

Combining like terms:

$\frac{2x^2 + 6x + 1}{(x-2)(x+1)} = \frac{21}{(x-2)(x+1)}$

Since the denominators are the same, we can equate the numerators:

$2x^2 + 6x + 1 = 21$

Rearranging the equation:

$2x^2 + 6x + 1 - 21 = 0$

$2x^2 + 6x - 20 = 0$

Dividing the equation by 2 to simplify:

$x^2 + 3x - 10 = 0$

Now, we can factor the quadratic equation:

$(x+5)(x-2) = 0$

So the possible solutions are:

$x + 5 = 0$ which gives $x = -5$

$x - 2 = 0$ which gives $x = 2$

Therefore, the solutions to the given equation are $x = -5$ and $x = 2$.