Question

23.1)(x-2)^2-(3-x)^2=0 2)(2x+1)^2-(4-2x)^2=0 3)3(2x-5)^2-4(3-2x)^2+4x^2=0 4)0,2(3x-1,5)^2-0,4(1-2x)^2-0,2x^2=0​

Answer 1

Ответ:

Смотри решение на фото выше

Answer 2

Let's solve each of these quadratic equations one by one:

1. $(x-2)^2 - (3-x)^2 = 0$

To solve this equation, expand the square terms and simplify:

$$(x^2 - 4x + 4) - ((9 - 6x + x^2)) = 0$$

Now, combine like terms:

$$x^2 - 4x + 4 - 9 + 6x - x^2 = 0$$

Simplify further:

$$-5 = 0$$

The equation $(x-2)^2 - (3-x)^2 = 0$ has no real solutions because it simplifies to -5 = 0, which is not a true statement. Therefore, this equation has no solutions.

2. $(2x+1)^2 - (4-2x)^2 = 0$

Expand and simplify:

$$(4x^2 + 4x + 1) - (16 - 16x + 4x^2) = 0$$

Combine like terms:

$$4x^2 + 4x + 1 - 16 + 16x - 4x^2 = 0$$

Simplify further:

$$4x - 15 = 0$$

Now, isolate x:

$$4x = 15$$

Divide both sides by 4:

$$x = \frac{15}{4}$$

So, the solution to the equation $(2x+1)^2 - (4-2x)^2 = 0$ is $x = \frac{15}{4}$.

3. $3(2x-5)^2 - 4(3-2x)^2 + 4x^2 = 0$

First, expand and simplify each term:

$$3(4x^2 - 20x + 25) - 4(9 - 12x + 4x^2) + 4x^2 = 0$$

Now, distribute the constants:

$$12x^2 - 60x + 75 - 36 + 48x - 16x^2 + 4x^2 = 0$$

Combine like terms:

$$12x^2 - 16x^2 + 4x^2 - 60x + 48x + 75 - 36 = 0$$

Simplify further:

$$(12 - 16 + 4)x^2 + (48 - 60)x + (75 - 36) = 0$$

Now, simplify the coefficients:

$$0x^2 - 12x + 39 = 0$$

This equation simplifies to:

$$-12x + 39 = 0$$

Isolate x:

$$-12x = -39$$

Divide by -12:

$$x = \frac{39}{12} = \frac{13}{4}$$

So, the solution to the equation $3(2x-5)^2 - 4(3-2x)^2 + 4x^2 = 0$ is $x = \frac{13}{4}$.

4. $0.2(3x-1.5)^2 - 0.4(1-2x)^2 - 0.2x^2 = 0$