Question

1) (х+3)² –9= 2) (х-5)² + 10х= 3) (х+2) × (х+4) –х² –8= 4) (х+8y) × (x-7y) – x (x+y)=

Answer 1

Ответ:

1)(x+3-3)•(x+3+3)=x•(x+6)

2) x^2-10x+25+10x=x^2+25

3) x^2+4x+2x+8-x^-8=6x

4) x^2-7xy+8xy-56y^2-x^2-xy=0-56y^2=-56y^2

Объяснение:

^ обозначение степени

Answer 2

Let's solve each of these quadratic equations:

1. (x+3)² - 9 = 0

First, expand the left side of the equation:

(x+3)² - 9 = x² + 6x + 9 - 9

Now, simplify the equation:

x² + 6x = 0

To solve for x, you can factor out x:

x(x + 6) = 0

Now, set each factor equal to zero:

x = 0 or x + 6 = 0

For the second equation:

2. (x-5)² + 10x = 0

First, expand the left side of the equation:

(x-5)² + 10x = x² - 10x + 25 + 10x

Now, simplify the equation:

x² + 25 = 0

Subtract 25 from both sides of the equation:

x² = -25

To solve for x, take the square root of both sides. Note that since you have a negative number under the square root, there are no real solutions for x in this equation.

For the third equation:

3. (x+2) * (x+4) - x² - 8 = 0

First, expand the left side of the equation:

(x+2)(x+4) - x² - 8 = x² + 4x + 2x + 8 - x² - 8

Now, simplify the equation:

6x = 0

Divide both sides by 6:

x = 0

For the fourth equation:

4. (x+8y) * (x-7y) - x(x+y) = 0

First, expand the left side of the equation:

(x+8y)(x-7y) - x(x+y) = x² - 7xy + 8xy - 56y² - x² - xy

Now, simplify the equation:

x² - 56y² = 0

To solve for x, isolate the x² term:

x² = 56y²

Take the square root of both sides:

x = ±√(56y²)

x = ±2√14y

So, the solutions for each equation are as follows:

1. x = 0 or x = -6
2. No real solutions
3. x = 0
4. x = 2√14y or x = -2√14y