Question

Ab+c²=13(a-c)(b-c)+(a+c)(b+c)=​

Answer 1

Ответ:

аб-ас-сб+с^2+аб+ас+сб+с^2=2с^2+2аб=2×(с^2+аб) *подставляем*

2×13=26

Answer 2

To simplify the expression $ab+c^2=13(a-c)(b-c)+(a+c)(b+c)$, let's start by expanding and simplifying both sides:

Left side: $ab + c^2$

Right side: $13(a-c)(b-c) + (a+c)(b+c)$

Now, let's expand the right side:

$13(a-c)(b-c) + (a+c)(b+c)$ $13(ab - ac - bc + c^2) + (ab + ac + bc + c^2)$

Now, distribute the terms on both sides:

$13ab - 13ac - 13bc + 13c^2 + ab + ac + bc + c^2$

Now, combine like terms:

$(13ab + ab) + (-13ac + ac) + (-13bc + bc) + (13c^2 + c^2)$

Simplify each group of terms:

$14ab - 13ac - 13bc + 14c^2$

So, after simplifying both sides of the equation, you have:

$ab + c^2 = 14ab - 13ac - 13bc + 14c^2$

Please note that I've simplified the equation according to basic algebraic rules. If you have specific values for $a$ and $b$, you can substitute them in to further simplify the equation.