Question

X+2/x-1+x+3/x+1=x+5/x^2-1

Answer 1

Объяснение:

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Answer 2

To solve the equation $\frac{x+2}{x-1} + \frac{x+3}{x+1} = \frac{x+5}{x^2-1}$, we'll first find a common denominator and then simplify the equation.

The common denominator for the fractions on the left side of the equation will be $(x-1)(x+1) = x^2 - 1$. So, we'll rewrite each fraction with this common denominator:

$\frac{(x+2)(x+1)}{(x-1)(x+1)} + \frac{(x+3)(x-1)}{(x+1)(x-1)} = \frac{x+5}{x^2-1}$

Now, the denominators are the same, so we can combine the fractions on the left side of the equation:

$\frac{(x+2)(x+1) + (x+3)(x-1)}{x^2 - 1} = \frac{x+5}{x^2-1}$

Next, we can simplify the numerators:

$(x+2)(x+1) + (x+3)(x-1) = x+5$

Now, expand and simplify the left side of the equation:

$x^2 + 3x + 2 + x^2 + 2x - 3 = x + 5$

Combine like terms:

$2x^2 + 5x - 1 = x + 5$

Now, we'll move all terms to one side of the equation:

$2x^2 + 5x - 1 - x - 5 = 0$

Simplify the equation:

$2x^2 + 4x - 6 = 0$

Divide the entire equation by 2 to simplify further:

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

Now, let's factor this quadratic equation:

$(x + 3)(x - 1) = 0$

Set each factor equal to zero and solve for $x$:

1. $x + 3 = 0$: $x = -3$

2. $x - 1 = 0$: $x = 1$

So, the solutions to the equation are $x = -3$ and $x = 1$.