Question

(Х+1)(Х-1) - (Х-2) (х+3)=0

Answer 1

Объяснение:

(x+1)(x-1)-(x-2)(x+3)=0

раскрываем скобки

x² -x +x -1 (-x+2)(x+3)=0

x²-x+x-1-x -3x +2x +5=0

взаимоуничожаем и получаем

x²-x-3x+2x=-4

x²-2x=-4

x=5

Answer 2

To solve the quadratic equation $(x+1)(x-1) - (x-2)(x+3) = 0$, you can follow these steps:

1. Expand and simplify the equation:

$(x+1)(x-1) - (x-2)(x+3) = x^2 - 1 - (x^2 - 5x + 6) = x^2 - 1 - x^2 + 5x - 6$

Now, simplify further:

$-1 + 5x - 6 = 5x - 7$

2. Set the simplified equation equal to 0:

$5x - 7 = 0$

3. Add 7 to both sides of the equation to isolate the variable:

$5x = 7$

4. Finally, divide both sides by 5 to solve for x:

$x = \frac{7}{5}$

So, the solution to the equation $(x+1)(x-1) - (x-2)(x+3) = 0$ is $x = \frac{7}{5}$.