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\(\hept{\begin{cases}a^2+ab+\frac{b^2}{3}=25\\c^2+\frac{b^2}{3}=9\end{cases}}\Rightarrow a^2+ac-c^2=16\)
\(\Rightarrow a^2+ab-c^2=a^2+ac+c^2\left(=16\right)\)
\(\Rightarrow ab-c^2=ac+c^2\)
\(\Rightarrow ab=ac+2c^2\)
\(\Rightarrow ab+ac=2ac+2c^2\)
\(\Leftrightarrow a\left(b+c\right)=2c\left(a+c\right)\)
\(\Leftrightarrow\frac{2c}{a}=\frac{b+c}{a+c}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}a^2+ab+\dfrac{b^2}{3}=25\\c^2+\dfrac{b^2}{3}=9\\a^2+ac+c^2=16\end{matrix}\right.\)
\(\Rightarrow a^2+ab+\dfrac{b^2}{3}=c^2+\dfrac{b^2}{3}+a^2+ac+c^2\)
\(\Rightarrow ab=2c^2+ac\)
Biến đổi 1 chút là ra
\(\rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có \(a^2+ab+\frac{b^2}{3}=c^2+\frac{b^2}{3}+a^2+ac+c^2\left(=25\right)\)
\(\Rightarrow a^2+ab+\frac{b^2}{3}=2c^2+\frac{b^2}{3}+a^2+ac\\ \Rightarrow ab=2c^2+ac\\ \Rightarrow ab+ac=2c^2+2ac\\ \Rightarrow a\left(b+c\right)=2c\left(a+c\right)\\ \Rightarrow\frac{2c}{a}=\frac{b+c}{a+c}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(a^2+ab+\dfrac{b^2}{3}=c^2+\dfrac{b^2}{3}+a^2+ac+c^2\)
\(\Rightarrow a^2+ab+\dfrac{b^2}{3}=2c^2+\dfrac{b^2}{3}+a^2+ac\)
\(\Rightarrow ab=2c^2+ac\)
\(\Rightarrow ab+ac=2ac+2c^2\)
\(\Rightarrow a\left(b+c\right)=2c\left(a+c\right)\)
\(\Rightarrow\dfrac{2c}{a}=\dfrac{b+c}{a+c}\left(đpcm\right)\)