Question

А)x^2-5/x-3=4x/x-3б)x-2/x+1-x-4/x+2=0​

Answer 1

а) вроде так. б не знаю как делать

Answer 2

Let's solve each equation step by step:

A) $\frac{x^2 - 5}{x - 3} = \frac{4x}{x - 3}$

To solve this equation, we first need to get rid of the denominators by multiplying both sides of the equation by $(x - 3)$:

$(x - 3) \cdot \frac{x^2 - 5}{x - 3} = (x - 3) \cdot \frac{4x}{x - 3}$

Simplify the equation:

$x^2 - 5 = 4x$

Now, let's move all terms to one side of the equation to set it to zero:

$x^2 - 4x - 5 = 0$

Now, we have a quadratic equation. We can solve it by factoring or using the quadratic formula. Factoring is the easiest in this case:

$(x - 5)(x + 1) = 0$

Now, we set each factor to zero and solve for $x$:

$x - 5 = 0$ or $x + 1 = 0$

$x = 5$ or $x = -1$

So, the solutions to the equation are $x = 5$ and $x = -1$.

B) $\frac{x - 2}{x + 1} - \frac{x - 4}{x + 2} = 0$

To solve this equation, we first need to find a common denominator for the fractions. The common denominator is $(x + 1)(x + 2)$. We then adjust the fractions:

$\frac{(x - 2)(x + 2)}{(x + 1)(x + 2)} - \frac{(x - 4)(x + 1)}{(x + 1)(x + 2)} = 0$

Now, combine the fractions:

$\frac{(x - 2)(x + 2) - (x - 4)(x + 1)}{(x + 1)(x + 2)} = 0$

Expand the expressions in the numerator:

$\frac{x^2 - 2x + 2x - 4 - (x^2 - 4x + x - 4)}{(x + 1)(x + 2)} = 0$

Simplify the numerator:

$\frac{x^2 - 4 - (x^2 - 3x - 4)}{(x + 1)(x + 2)} = 0$

Now, continue simplifying the equation:

$\frac{x^2 - 4 - x^2 + 3x + 4}{(x + 1)(x + 2)} = 0$

Combine like terms in the numerator:

$\frac{3x}{(x + 1)(x + 2)} = 0$

Since the fraction is equal to zero, the numerator must be zero:

$3x = 0$

Now, solve for $x$:

$x = 0$

So, the solution to the equation is $x = 0$.