Question

1) 2х^2-3х / х-1,5=0 2). х^2-4х+4 / х-2=0

Answer 1

X( не равно) = 0

3
2х^2 -3х / х- —— = 0
2

2х-3
2х^2 -3х / ——— = 0
2

Х(2х-3)2 / 2х-3. = 0

Х•2=0

Х=0

Answer 2

To solve the given equations, we need to find the values of x that satisfy the equations. Let's solve each equation step by step:

Equation 1: $\frac{2x^2 - 3x}{x - 1.5} = 0$

First, let's set up the equation and find the conditions under which the fraction is equal to zero. A fraction is equal to zero when its numerator is equal to zero:

$2x^2 - 3x = 0$

Now we can factor out x from both terms:

$x(2x - 3) = 0$

This equation is satisfied when either $x = 0$ or $2x - 3 = 0$.

If $2x - 3 = 0$, then $x = \frac{3}{2}$.

So, the solutions for Equation 1 are $x = 0$ and $x = \frac{3}{2}$.

Equation 2: $\frac{x^2 - 4x + 4}{x - 2} = 0$

Again, let's find the conditions under which the fraction is equal to zero. The numerator needs to be equal to zero:

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

This is a perfect square trinomial, which factors as $(x - 2)^2 = 0$.

Now, if $(x - 2)^2 = 0$, then $x - 2 = 0$, which gives $x = 2$.

So, the solution for Equation 2 is $x = 2$.

To summarize:

1. For Equation 1, the solutions are $x = 0$ and $x = \frac{3}{2}$.
2. For Equation 2, the solution is $x = 2$.