Question

(a+b)+(b+c)+(a+c) если a+b+c=49

Answer 1

Посуди логически от перемены мест слагаемых сумма не меняется значит:49+49=98

Answer 2

I can understand and communicate fluently in your language of choice. You asked me to answer in detail. $(a+b)+(b+c)+(a+c)$ if $a+b+c=49$.

There are different ways to approach this problem, but one possible method is to use the fact that $(a+b)+(b+c)+(a+c) = 2(a+b+c) - (a^2+b^2+c^2)$. Then, we can substitute $a+b+c=49$ and simplify the expression. Here is a step-by-step solution:

\begin{align*} (a+b)+(b+c)+(a+c) &= 2(a+b+c) - (a^2+b^2+c^2) \\ &= 2(49) - (a^2+b^2+c^2) \\ &= 98 - (a^2+b^2+c^2) \end{align*}

To find the value of $(a^2+b^2+c^2)$, we can use another identity: $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$. Then, we can substitute $a+b+c=49$ and solve for $(a^2+b^2+c^2)$. Here is how:

\begin{align*} (a+b+c)^2 &= a^2+b^2+c^2+2(ab+bc+ca) \\ 49^2 &= a^2+b^2+c^2+2(ab+bc+ca) \\ 2401 &= a^2+b^2+c^2+2(ab+bc+ca) \\ a^2+b^2+c^2 &= 2401 - 2(ab+bc+ca) \end{align*}

Now, we can plug this value into the original expression and get the final answer:

\begin{align*} (a+b)+(b+c)+(a+c) &= 98 - (a^2+b^2+c^2) \\ &= 98 - (2401 - 2(ab+bc+ca)) \\ &= 98 + 2(ab+bc+ca) - 2401 \\ &= 2(ab+bc+ca) - 2303 \end{align*}

Therefore, $(a+b)+(b+c)+(a+c) = 2(ab+bc+ca) - 2303$ if $a+b+c=49$. I hope this helps you understand the problem better. If you have any other questions, feel free to ask me.Ответь+подробно.+(a+b)+(b+c)+(a+c)+если+a+b+c=49