Question

(x-5)(x+5)=(x+3) квадрат​

Answer 1

Ответ:

Пошаговое объяснение: вопросы в комм

Answer 2

х²-5²=(х+3)

х²+25-(х+3)=0

х²+25-х-3=0

х²-х+22=0

D=(-1)-4×1×22=1-88= -87<0

Уравнение не имеет решения

Answer 3

To solve the equation $(x-5)(x+5) = (x+3)^2$, we'll start by expanding both sides of the equation and then simplifying to find the value(s) of $x$ that satisfy the equation.

Expanding the left side: $(x-5)(x+5) = x^2 + 5x - 5x - 25 = x^2 - 25$

Expanding the right side: $(x+3)^2 = x^2 + 6x + 9$

So, the equation becomes: $x^2 - 25 = x^2 + 6x + 9$

Now, we can subtract $x^2$ from both sides of the equation to simplify: $-25 = 6x + 9$

Next, subtract 9 from both sides: $-34 = 6x$

Finally, divide both sides by 6 to solve for $x$: $x = -\frac{34}{6} = -\frac{17}{3}$

So, the solution for the equation $(x-5)(x+5) = (x+3)^2$ is $x = -\frac{17}{3}$.