Question

[nucuoicuapi]

Cho

\(\left\{{}\begin{matrix}a+b+c=5\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\end{matrix}\right.\)

Tính

\(a^2+b^2+c^2\)

Answer 1

[Mashiro Shiina]

Đề ảo tek.Sửa đề.

\(\left\{{}\begin{matrix}a+b+c=5\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a+b+c\right)^2=25\\\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+c^2+2ab+2bc+2ac=25\\bc+ac+ab=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+c^2+2ab+2bc+2ac=25\\2bc+2ac+2ab=0\end{matrix}\right.\)

\(\Leftrightarrow a^2+b^2+c^2+2ab-2ab+2bc-2bc+2ac-2ac=25\)

\(\Leftrightarrow a^2+b^2+c^2=25\)