Question

7) (7х – 14)(2х + 5) = 0; 8) 2(х + 1)(6 – 3х) = 0; 9) х2 – 16 = 0;

Answer 1

7) 7х - 14=0 или 2х+5=0

х = 2

x = -2,5

8) x+1=0 или 6-3х=0

х=-1

х=2

9)х²=16

х=4

х=-4

Answer 2

Відповідь:

7) x = 2

x = -2,5

8) х = -1

x = 2

9) х = 4

х = -4

Пояснення:

7) (7х – 14)(2х + 5) = 0

7х - 14=0

7x = 14

x = 14 : 7

х = 2

2х + 5 = 0

2x = -5

x = -5 : 2

x = -2,5

8) 2(х + 1)(6 – 3х) = 0

x+1=0

х = -1

6 - 3х = 0

-3x = -6

x = -6 : (-3)

х = 2

9) х² – 16 = 0

х² = 16

х = 4

х = -4

Answer 3

It seems like you've provided three equations, and it looks like you're interested in finding the solutions for the variable x in each equation. Let's solve them one by one:

1. $(7x - 14)(2x + 5) = 0$

To solve this equation, we need to set each factor equal to zero and solve for x:

First factor: $7x - 14 = 0$ $7x = 14$ $x = 2$

Second factor: $2x + 5 = 0$ $2x = -5$ $x = -\frac{5}{2}$

So, the solutions for this equation are $x = 2$ and $x = -\frac{5}{2}$.

2. $2(x + 1)(6 - 3x) = 0$

Again, set each factor equal to zero and solve for x:

First factor: $x + 1 = 0$ $x = -1$

Second factor: $6 - 3x = 0$ $-3x = -6$ $x = 2$

So, the solutions for this equation are $x = -1$ and $x = 2$.

3. $x^2 - 16 = 0$

This is a quadratic equation that can be factored as the difference of squares:

$x^2 - 16 = (x + 4)(x - 4) = 0$

Setting each factor equal to zero:

First factor: $x + 4 = 0$ $x = -4$

Second factor: $x - 4 = 0$ $x = 4$

The solutions for this equation are $x = -4$ and $x = 4$.

So, the solutions for the three equations are as follows:

1. $x = 2$ and $x = -\frac{5}{2}$
2. $x = -1$ and $x = 2$
3. $x = -4$ and $x = 4$