CMR nếu (a+b+c)^2 = 3(ab+bc+ca) thì a=b=c 22/07/2021 Bởi Ariana CMR nếu (a+b+c)^2 = 3(ab+bc+ca) thì a=b=c
Ta có: $(a + b + c)^2 = 3(ab +bc + ca)$ $\Leftrightarrow a^2 + b^2 + c^2 + 2(ab + bc + ca) = 3(ab + bc + ca)$ $\Leftrightarrow a^2 +b^2 + c^2 – (ab + bc + ca) = 0$ $\Leftrightarrow 2(a^2 + b^2 + c^2) – 2(ab + bc + ca) = 0$ $\Leftrightarrow (a^2 – 2ab + b^2) + (b^2 – 2bc + c^2) + (c^2 – 2ca + a^2) = 0$ $\Leftrightarrow (a -b)^2 + (b-c)^2 + (c -a)^2 = 0$ $\Leftrightarrow \begin{cases}a – b = 0\\b – c = 0\\c – a = 0\end{cases}\Leftrightarrow a = b = c$ Bình luận
Ta có:
$(a + b + c)^2 = 3(ab +bc + ca)$
$\Leftrightarrow a^2 + b^2 + c^2 + 2(ab + bc + ca) = 3(ab + bc + ca)$
$\Leftrightarrow a^2 +b^2 + c^2 – (ab + bc + ca) = 0$
$\Leftrightarrow 2(a^2 + b^2 + c^2) – 2(ab + bc + ca) = 0$
$\Leftrightarrow (a^2 – 2ab + b^2) + (b^2 – 2bc + c^2) + (c^2 – 2ca + a^2) = 0$
$\Leftrightarrow (a -b)^2 + (b-c)^2 + (c -a)^2 = 0$
$\Leftrightarrow \begin{cases}a – b = 0\\b – c = 0\\c – a = 0\end{cases}\Leftrightarrow a = b = c$