Giải:
Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)
Ta có: \(a^2+3b^2-2c^2=-16\)
\(\Rightarrow4k^2+27k^2-32k^2=-16\)
\(\Rightarrow-k^2=-16\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=\pm4\)
+) \(k=4\Rightarrow a=8,b=12,c=16\)
+) \(k=-4\Rightarrow a=-8;b=-12;c=-16\)
Vậy bộ số \(\left(a;b;c\right)\) là \(\left(8;12;16\right);\left(-8;-12;-16\right)\)
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a^2}{4}=\dfrac{3b^2}{27}=\dfrac{2c^2}{32}=\dfrac{a^2+3b^2-2c^2}{4+27-32}=\dfrac{-16}{-1}=16\)
\(\Rightarrow a^2=64,b^2=144,c^2=256\) hay:
\(\left(a;b;c\right)=\left(8;12;16\right)=\left(-8;-12;-16\right)\)
ĐS: \(\left(a;b;c\right)=\left(8;12;16\right)=\left(-8;-12;-16\right)\)
Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)
Ta có: \(a^2+3b^2-2c^2=-16\)
\(\Leftrightarrow\left(2k\right)^2+3\left(3k\right)^2-2\left(4k\right)^2=-16\)
\(\Leftrightarrow4k^2+3\cdot9k^2-2\cdot16k^2=-16\)
\(\Leftrightarrow4k^2+27k^2-32k^2=-16\)
\(\Leftrightarrow\left(4+27-32\right)k^2=-16\)
\(\Leftrightarrow-k^2=-16\)
\(\Leftrightarrow k^2=16\)
\(\Leftrightarrow k=\sqrt{16}\)
\(\Leftrightarrow k=\pm4\)
Nếu \(k=-4\) thì:
\(a=2\left(-4\right)=-8\)
\(b=3\left(-4\right)=-12\)
\(c=4\left(-4\right)=-16\)
Nếu \(k=4\) thì :
\(a=2\cdot4=8\)
\(b=3\cdot4=12\)
\(c=4\cdot4=16\)
Vậy \(a;b;c=\left\{-8;-12;-16\right\}\) hoặc \(a;b;c=\left\{8;12;16\right\}\)