\(a,\) Áp dụng tc dtsbn:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{x-1-2\left(y-2\right)+3\left(z-3\right)}{2-2\cdot3+3\cdot4}=\dfrac{\left(x-2y+3z\right)-1+4-9}{2-6+12}=\dfrac{8}{8}=1\\ \Rightarrow\left\{{}\begin{matrix}x-1=2\\y-2=3\\z-3=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)
\(b,\) Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\Rightarrow x=3k;y=4k;z=5k\)
\(2x^2+2y^2-3z^2=-100\\ \Rightarrow18k^2+32k^2-75k^2=-100\\ \Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=6;y=8;z=10\\x=-6;y=-8;z=-10\end{matrix}\right.\)
mai giang phải nộp bài rùi ạ giúp với đi ạ hứa sẽ like mà
