Có \(a^2+ab+\frac{b^2}{3}=c^2+\frac{b^2}{3}+a^2+ac+c^2\left(=25\right)\)

\(\Rightarrow a^2+ab+\frac{b^2}{3}=2c^2+\frac{b^2}{3}+a^2+ac\\ \Rightarrow ab=2c^2+ac\\ \Rightarrow ab+ac=2c^2+2ac\\ \Rightarrow a\left(b+c\right)=2c\left(a+c\right)\\ \Rightarrow\frac{2c}{a}=\frac{b+c}{a+c}\)

\(\Leftrightarrow a^2+ab+\frac{b^2}{3}=c^2+\frac{b^2}{3}+a^2+ac+c^2\)

\(\Leftrightarrow ab=2c^2+ca\Leftrightarrow ab+ac=2c^2+2ac\)

\(\Leftrightarrow a\left(b+c\right)=2c\left(a+c\right)\Rightarrow\frac{2c}{a}=\frac{b+c}{a+c}\rightarrowđpcm\)

\(\left\{{}\begin{matrix}a^2+ab+\dfrac{b^2}{3}=25\\c^2+\dfrac{b^2}{3}=9\\a^2+ac+c^2=16\end{matrix}\right.\)

\(\Rightarrow a^2+ab+\dfrac{b^2}{3}=c^2+\dfrac{b^2}{3}+a^2+ac+c^2\)

\(\Rightarrow ab=2c^2+ac\)

Biến đổi 1 chút là ra

\(\rightarrowđpcm\)

\(\hept{\begin{cases}a^2+ab+\frac{b^2}{3}=25\\c^2+\frac{b^2}{3}=9\end{cases}}\Rightarrow a^2+ac-c^2=16\)

\(\Rightarrow a^2+ab-c^2=a^2+ac+c^2\left(=16\right)\)

\(\Rightarrow ab-c^2=ac+c^2\)

\(\Rightarrow ab=ac+2c^2\)

\(\Rightarrow ab+ac=2ac+2c^2\)

\(\Leftrightarrow a\left(b+c\right)=2c\left(a+c\right)\)

\(\Leftrightarrow\frac{2c}{a}=\frac{b+c}{a+c}\left(đpcm\right)\)

Bài 2/a 

Giả sử \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\Rightarrow\hept{\begin{cases}a=2k\\b=3k\\c=5k\end{cases}}\)

\(\Rightarrow\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}\)

\(\Rightarrow\frac{3\cdot2k-2\cdot3k}{5}=\frac{2\cdot5k-5\cdot2k}{3}=\frac{5\cdot3k-3\cdot5k}{2}\)

\(\Rightarrow\frac{6k-6k}{5}=\frac{10k-10k}{3}=\frac{15k-15k}{2}\)

\(\Rightarrow\frac{0}{5}=\frac{0}{3}=\frac{0}{2}=0\left(đpcm\right)\)

Bài 2/c

Có a = 2k ; b = 3k ; c = 5k

=> 2 (a - b) (b - c) = a²

=> 2 (2k - 3k) (3k - 5k) = (2k)²

=> 2 (-1)k . (-2)k = 4k²

=> 4k² = 4k² (đpcm)

Mình chỉ làm được có vậy thôi, mong bạn thông cảm =))

Chúc bạn học tốt =))

\(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}\)

\(\Rightarrow\frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

 \(\frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}=\frac{15a-10b+6c-15a+10b-6c}{25+9+4}=0\)

\(\Rightarrow\hept{\begin{cases}\frac{15a-10b}{25}=0\\\frac{6c-15a}{9}=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}3a-2b=0\\2c-5a=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}3a=2b\\2c=5a\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\frac{a}{2}=\frac{b}{3}\\\frac{c}{5}=\frac{a}{2}\end{cases}}\)

                                                                                                                   \(\Rightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{5}\)