Question

Сократи дробь (x−5)2^\x^2+2x−35

Answer 1

$x^2+2x-35=0\; \; \to \; \; x_1=-7\; ,\; x_2=5\; \; \Rightarrow \; \; x^2+2x-35=(x+7)(x-5)\\\\\\\frac{(x-5)^2}{x^2+2x-35}=\frac{(x-5)^2}{(x+7)(x-5)}=\frac{x-5}{x+7}$

Answer 2

Ответ:

Х-5^2/х^2+7х-5х-35

Х-5^2/х(х+7)-5(х+7)

Х-5^2/(х+7)(х-5)

Х-5/х+7

Answer 3

To simplify the given fraction, you need to factorize the quadratic expression in the denominator if possible. Let's start by factoring the denominator:

The denominator is $x^2 + 2x - 35$, which is a quadratic expression.

Step 1: Find two numbers whose product is -35 and whose sum is 2 (the coefficient of the x term). The numbers are 7 and -5.

Step 2: Rewrite the middle term (2x) using the two numbers found in step 1: $x^2 + 7x - 5x - 35$.

Step 3: Group the terms and factor them: $x(x + 7) - 5(x + 7)$.

Step 4: Factor out the common factor (x + 7): $(x + 7)(x - 5)$.

Now that we have factored the denominator, the fraction becomes:

$\frac{(x - 5)^2}{(x + 7)(x - 5)}.$

Notice that there is a common factor of (x - 5) in both the numerator and the denominator. We can cancel out this common factor:

$\frac{\cancel{(x - 5)}(x - 5)}{(x + 7)\cancel{(x - 5)}}.$

The remaining fraction is:

$\frac{x - 5}{x + 7}.$

So, the simplified form of the fraction is $\frac{x - 5}{x + 7}$.