Question

1. (2х+1)²-4х²=3х+2 2. (3х+2)²-9х²=9х+7 3. (х-1)²-2(х+1)+1=0 4. (х-2)²-4(х-2)-5=0

Answer 1

Объяснение:

Вот решение уравнений на фотографиях

Answer 2

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

1. $(2x+1)^2 - 4x^2 = 3x+2$

First, expand the left side:

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

Now, the equation becomes:

$4x + 1 - 4x^2 = 3x + 2$

Subtract $3x + 2$ from both sides:

$4x - 3x + 1 - 4x^2 - 2 = 0$

Simplify:

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

Rearrange:

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

This is a quadratic equation. You can solve it using the quadratic formula:

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

In this case, $a = -4$, $b = 1$, and $c = -1$. Plug these values into the formula:

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

$x = \frac{-1 \pm \sqrt{1 - 16}}{-8}$

$x = \frac{-1 \pm \sqrt{-15}}{-8}$

Since the discriminant ($1 - 16 = -15$) is negative, this equation has no real solutions.

2. $(3x+2)^2 - 9x^2 = 9x+7$

Expand the left side:

$(3x+2)^2 = (3x+2)(3x+2) = 9x^2 + 12x + 4$

Now, the equation becomes:

$9x^2 + 12x + 4 - 9x^2 = 9x + 7$

Subtract $9x^2$ from both sides:

$12x + 4 = 9x + 7$

Subtract $9x$ from both sides:

$3x + 4 = 7$

Subtract 4 from both sides:

$3x = 3$

Divide by 3:

$x = 1$

So, the solution to this equation is $x = 1$.

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

Expand the left side:

$(x-1)^2 = (x-1)(x-1) = x^2 - 2x + 1$

Now, the equation becomes:

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

Distribute the -2 on the second term:

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

Combine like terms:

$x^2 - 4x = 0$

Factor out an x:

$x(x - 4) = 0$