Question

(x-7)(x-2)+(x+4)(x-2)=(x-2)(x+2)

Answer 1

(х - 7)(х - 2) + (х + 4)(х - 2) = (х - 2)(х + 2)

(х - 2) · ((х - 7) + (х + 4)) = х² - 2²

(х - 2) · (2х - 3) = х² - 4

2х² - 4х - 3х + 6 = х² - 4

2х² - 7х + 6 - х² + 4 = 0

х² - 7х + 10 = 0

D = b² - 4ac = (-7)² - 4 · 1 · 10 = 49 - 40 = 9

√D = √9 = 3

х₁ = (7-3)/(2·1) = 4/2 = 2

х₂ = (7+3)/(2·1) = 10/2 = 5

Ответ: (2; 5).

Проверка:

(2 - 7)(2 - 2) + (2 + 4)(2 - 2) = (2 - 2)(2 + 2)

-5 · 0 + 6 · 0 = 0 · 4

0 = 0 - верно

(5 - 7)(5 - 2) + (5 + 4)(5 - 2) = (5 - 2)(5 + 2)

-2 · 3 + 9 · 3 = 3 · 7

-6 + 27 = 21

21 = 21 - верно

Answer 2

(x-7)(x-2)+(x+4)(x-2)=(x-2)(x+2)

(х-7+х+4)(х-2)=(х+2)(х-2)

х-2=0

х=2

Сократим на (х-2)

х-7+х+4 = х+2

2х-х=2+3

х=5

Ответ: 2, 5

Answer 3

To solve the given equation $(x-7)(x-2) + (x+4)(x-2) = (x-2)(x+2)$, let's first simplify both sides of the equation and then solve for $x$.

Expanding each expression:

Left side: $(x-7)(x-2) + (x+4)(x-2)$ $= x^2 - 9x + 14 + x^2 + 2x - 8$ $= 2x^2 - 7x + 6$

Right side: $(x-2)(x+2)$ $= x^2 - 4$

Now, we can set the left side equal to the right side and solve for $x$:

$2x^2 - 7x + 6 = x^2 - 4$

First, let's move all terms to one side to get a quadratic equation:

$2x^2 - 7x + 6 - x^2 + 4 = 0$

Simplifying:

$x^2 - 7x + 10 = 0$

Now, let's factor the quadratic equation:

$(x - 5)(x - 2) = 0$

This equation is satisfied when $x - 5 = 0$ or $x - 2 = 0$. Therefore, the solutions for $x$ are:

$x_1 = 5$ and $x_2 = 2$.