Question

1) 12x²-2÷9=0 2) 3,7x-x²=0 3) 12x²=0 4) x²-3,7x=0

Answer 1

Пошаговое объяснение:

Answer 2

It seems like you've provided a list of quadratic equations. Quadratic equations are in the form of $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants. To solve these equations, you can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Let's apply the quadratic formula to each of the equations you've provided:

1. $12x^2 - \frac{2}{9} = 0$: Here, $a = 12$, $b = 0$, and $c = -\frac{2}{9}$. Plugging these values into the quadratic formula:

   $x = \frac{-0 \pm \sqrt{0^2 - 4 \cdot 12 \cdot \left(-\frac{2}{9}\right)}}{2 \cdot 12}$ $x = \pm \frac{\sqrt{\frac{8}{9}}}{24} = \pm \frac{1}{6}\sqrt{\frac{2}{3}}$

2. $3.7x - x^2 = 0$: Here, $a = -1$, $b = 3.7$, and $c = 0$. Plugging these values into the quadratic formula:

   $x = \frac{-3.7 \pm \sqrt{3.7^2 - 4 \cdot (-1) \cdot 0}}{2 \cdot -1}$ $x = \frac{-3.7 \pm \sqrt{13.69}}{-2}$ $x = \frac{-3.7 \pm 3.7}{-2}$

   This gives two solutions: $x = 0$ and $x = 3.7$.

3. $12x^2 = 0$: Here, $a = 12$, $b = 0$, and $c = 0$. Plugging these values into the quadratic formula:

   $x = \frac{-0 \pm \sqrt{0^2 - 4 \cdot 12 \cdot 0}}{2 \cdot 12}$ $x = \frac{0}{24} = 0$

4. $x^2 - 3.7x = 0$: Here, $a = 1$, $b = -3.7$, and $c = 0$. Plugging these values into the quadratic formula:

   $x = \frac{3.7 \pm \sqrt{(-3.7)^2 - 4 \cdot 1 \cdot 0}}{2 \cdot 1}$ $x = \frac{3.7 \pm \sqrt{13.69}}{2}$ $x = \frac{3.7 \pm 3.7}{2}$

   This gives two solutions: $x = 0$ and $x = 3.7$.

So, for the equations you provided:

1. $x = \pm \frac{1}{6}\sqrt{\frac{2}{3}}$