Question

X^2-10x+21/x^2-4=0 даю 15 баллов

Answer 1

Решение и ответ во вложении. желаю удачи!

Answer 2

To solve the equation $\frac{{x^2-10x+21}}{{x^2-4}}=0$, we first need to find the values of $x$ that satisfy the equation.

To simplify the equation, let's factor the numerator and the denominator:

$\frac{{x^2-10x+21}}{{x^2-4}} = \frac{{(x-7)(x-3)}}{{(x-2)(x+2)}} = 0$

Now, to make the fraction equal to zero, the numerator $(x-7)(x-3)$ must be equal to zero, and the denominator $(x-2)(x+2)$ must not be equal to zero.

So, set the numerator equal to zero and solve for $x$:

$x-7 = 0$ or $x-3 = 0$

Solving these equations:

$x = 7$ or $x = 3$

Next, check whether the denominator is not equal to zero for these values of $x$:

For $x = 7$:

$(x-2)(x+2) = (7-2)(7+2) = 5 \times 9 = 45$, which is not zero.

For $x = 3$:

$(x-2)(x+2) = (3-2)(3+2) = 1 \times 5 = 5$, which is not zero.

Both $x = 7$ and $x = 3$ satisfy the equation.

Therefore, the solutions to the equation $\frac{{x^2-10x+21}}{{x^2-4}}=0$ are $x = 7$ and $x = 3$.

Since you provided 15 points for the solution, I hope this answer is satisfactory. If you have any more questions or need further assistance, feel free to ask!