Câu 3:
\(\dfrac{x}{y}=\dfrac{2}{3}\)
=>\(\dfrac{x}{2}=\dfrac{y}{3}\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\)
=>x=2k; y=3k
xy=24
=>\(2k\cdot3k=24\)
=>\(6k^2=24\)
=>\(k^2=4\)
=>\(\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
TH1: k=2
=>\(\left\{{}\begin{matrix}x=2\cdot2=4\\y=3\cdot2=6\end{matrix}\right.\)
TH2: k=-2
=>\(\left\{{}\begin{matrix}x=2\cdot\left(-2\right)=-4\\y=3\cdot\left(-2\right)=-6\end{matrix}\right.\)
Câu 4:
a: \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{4}\)
=>\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{2}\)
Đặt \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{2}=k\)
=>x=4k; y=3k; z=2k
xyz=-24
=>\(4k\cdot3k\cdot2k=-24\)
=>\(24k^3=-24\)
=>\(k^3=-1\)
=>k=-1
=>\(\left\{{}\begin{matrix}x=4\cdot\left(-1\right)=-4\\y=3\cdot\left(-1\right)=-3\\z=2\cdot\left(-1\right)=-2\end{matrix}\right.\)
b: \(\dfrac{x}{4}=\dfrac{y}{8}=\dfrac{z}{16}\)
=>\(\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{4}\)
Đặt \(\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{4}=k\)
=>\(\left\{{}\begin{matrix}x=k\\y=2k\\z=4k\end{matrix}\right.\)
\(x^2-y^2=-60\)
=>\(k^2-\left(2k\right)^2=-60\)
=>\(-3k^2=-60\)
=>\(k^2=20\)
=>\(\left[{}\begin{matrix}k=2\sqrt{5}\\k=-2\sqrt{5}\end{matrix}\right.\)
TH1: \(k=2\sqrt{5}\)
=>\(\left\{{}\begin{matrix}x=k=2\sqrt{5}\\y=2k=2\cdot2\sqrt{5}=4\sqrt{5}\\z=4k=4\cdot2\sqrt{5}=8\sqrt{5}\end{matrix}\right.\)
TH2: \(k=-2\sqrt{5}\)
=>\(\left\{{}\begin{matrix}x=k=-2\sqrt{5}\\y=2k=-2\cdot2\sqrt{5}=-4\sqrt{5}\\z=4k=4\cdot\left(-2\sqrt{5}\right)=-8\sqrt{5}\end{matrix}\right.\)
c: \(\dfrac{x}{y}=\dfrac{7}{10}\)
=>\(\dfrac{x}{7}=\dfrac{y}{10}\left(1\right)\)
\(\dfrac{y}{z}=\dfrac{5}{8}\)
=>\(\dfrac{y}{5}=\dfrac{z}{8}\)
=>\(\dfrac{y}{10}=\dfrac{z}{16}\left(2\right)\)
Từ (1),(2) suy ra \(\dfrac{x}{7}=\dfrac{y}{10}=\dfrac{z}{16}\)
mà 2x-y+3z=104
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{7}=\dfrac{y}{10}=\dfrac{z}{16}=\dfrac{2x-y+3z}{2\cdot7-10+3\cdot16}=\dfrac{104}{52}=2\)
=>\(\left\{{}\begin{matrix}x=2\cdot7=14\\y=2\cdot10=20\\z=2\cdot16=32\end{matrix}\right.\)
