Question

(x-2)² (x-4)² (3x-8)² ——- + ——— = ———— 5 4 20

Answer 1

Ответ:

На фото, сорри за ошибку;)

Answer 2

To solve the equation $\frac{{(x-2)^2}}{{5}} + \frac{{(x-4)^2}}{{4}} = \frac{{(3x-8)^2}}{{20}}$, you can start by finding a common denominator. In this case, the common denominator is 20.

So, you need to rewrite each fraction with a denominator of 20:

$\frac{{4(x-2)^2}}{{20}} + \frac{{5(x-4)^2}}{{20}} = \frac{{(3x-8)^2}}{{20}}$

Now, you have the equation:

$\frac{{4(x-2)^2 + 5(x-4)^2}}{20} = \frac{{(3x-8)^2}}{20}$

Next, you can multiply both sides of the equation by 20 to eliminate the fractions:

$4(x-2)^2 + 5(x-4)^2 = (3x-8)^2$

Expand each term:

$4(x^2 - 4x + 4) + 5(x^2 - 8x + 16) = (9x^2 - 48x + 64)$

Now, distribute the coefficients on the left side of the equation:

$4x^2 - 16x + 16 + 5x^2 - 40x + 80 = 9x^2 - 48x + 64$

Combine like terms:

$9x^2 - 56x + 96 = 9x^2 - 48x + 64$

Subtract $9x^2$ from both sides to eliminate the $x^2$ terms:

$-56x + 96 = -48x + 64$

Now, add $48x$ to both sides to eliminate the $x$ terms:

$-8x + 96 = 64$

Subtract 96 from both sides:

$-8x = -32$

Finally, divide both sides by -8 to solve for x:

$x = 4$

So, the solution to the equation is $x = 4$.