Question

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4};a^2+3b^2-2c^2=-64\)

Giúp mk vs!!!

Answer 1

Xét \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\)

\(\Rightarrow a=2k;b=3k;c=4k\) (1)

Thay (1) vào \(a^2+3b^2-2c^2\)

\(\Rightarrow\)\(a^2+3b^2-2c^2\)\(=\left(2k\right)^2+3\left(3k\right)^2-2\left(4k\right)^2\)

\(=4k^2+27k^2-32k^2=-k^2=-64\)

\(\Rightarrow k^2=64\Rightarrow\left[{}\begin{matrix}k=-8\\k=8\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=\pm16\\b=\pm24\\c=\pm32\end{matrix}\right.\)