Question

(4-х^2)^2=(4х^2)(4+х^2)+2х^2 (х^2-4)

Answer 1

Очень трудный пример.
20 минут решала

Answer 2

To solve the equation $(4-x^2)^2 = (4x^2)(4+x^2) + 2x^2(x^2-4)$, we can start by expanding both sides and simplifying:

Left side: $(4 - x^2)^2 = (4 - x^2)(4 - x^2) = 16 - 8x^2 + x^4$

Right side: $4x^2(4+x^2) + 2x^2(x^2-4) = 16x^2 + 4x^4 + 2x^4 - 8x^2$

Now, the equation becomes:

$16 - 8x^2 + x^4 = 16x^2 + 4x^4 + 2x^4 - 8x^2$

Now, let's simplify further:

Combine like terms on both sides:

$16 - 8x^2 + x^4 = 16x^2 + 6x^4 - 8x^2$

Now, subtract $16$ from both sides to get all terms on one side:

$x^4 - 8x^2 + 16 = 6x^4 - 8x^2$

Now, we can simplify the equation further:

$x^4 - 6x^4 - 8x^2 + 8x^2 + 16 = 0$

Notice that the $8x^2$ and $-8x^2$ terms cancel out:

$-5x^4 + 16 = 0$

Now, add $5x^4$ to both sides:

$16 = 5x^4$

Finally, divide both sides by 5 to solve for $x^4$:

$x^4 = \frac{16}{5}$

Now, to find $x$, take the fourth root of both sides:

$x = \pm \sqrt[4]{\frac{16}{5}}$

So, $x$ can be either $x = \sqrt[4]{\frac{16}{5}}$ or $x = -\sqrt[4]{\frac{16}{5}}$.