Question

X-3/x+2 - 20/x^2-4 + 2 = 0

Answer 1

Вот, держи решение,можешь не нужно не писать

Answer 2

To solve the equation $\frac{X-3}{x+2} - \frac{20}{x^2-4} + 2 = 0$, we first need to simplify the equation and find a common denominator for the fractions. The common denominator for $x+2$ and $x^2-4$ is $(x+2)(x-2)$.

Step 1: Find the common denominator for the fractions: $\frac{X-3}{x+2} = \frac{(X-3)(x-2)}{(x+2)(x-2)}$ $\frac{20}{x^2-4} = \frac{20}{(x+2)(x-2)}$

Step 2: Rewrite the equation with the common denominator: $\frac{(X-3)(x-2)}{(x+2)(x-2)} - \frac{20}{(x+2)(x-2)} + 2 = 0$

Step 3: Combine the fractions: $\frac{(X-3)(x-2) - 20 + 2(x+2)(x-2)}{(x+2)(x-2)} = 0$

Step 4: Expand and simplify the numerator: $\frac{Xx - 2X - 3x + 6 - 20 + 2(x^2 - 4)}{(x+2)(x-2)} = 0$

Step 5: Combine like terms: $\frac{Xx - 2X - 3x + 6 + 2x^2 - 8}{(x+2)(x-2)} = 0$

Step 6: Combine the terms in the numerator: $\frac{2x^2 + Xx - 2X - 3x - 2}{(x+2)(x-2)} = 0$

Step 7: Factor the quadratic in the numerator: $\frac{(2x+1)(x-2) - (2x+1)}{(x+2)(x-2)} = 0$

Step 8: Cancel out common factors: $\frac{(2x+1)(x-2) - 1}{(x+2)(x-2)} = 0$

Step 9: Simplify the equation: $\frac{(2x+1)(x-2) - 1}{(x+2)(x-2)} = 0$

Step 10: Set the numerator equal to zero and solve for $x$: $(2x+1)(x-2) - 1 = 0$

Step 11: Expand and solve for $x$: $2x^2 - 3x - 3 = 0$

To solve this quadratic equation, you can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 2$, $b = -3$, and $c = -3$.

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

$x = \frac{3 \pm \sqrt{9 + 24}}{4}$

$x = \frac{3 \pm \sqrt{33}}{4}$

So, the two solutions for $x$ are:

$x = \frac{3 + \sqrt{33}}{4}$