a: \(-0,52:x=-9,36:16,38\)
=>\(\dfrac{0.52}{x}=\dfrac{9.36}{16.38}=\dfrac{4}{7}\)
=>\(x=0.52\cdot\dfrac{7}{4}=0,91\)
b: \(\dfrac{4\dfrac{1}{4}}{2\dfrac{7}{8}}=\dfrac{x}{1,61}\)
=>\(\dfrac{x}{1.61}=\dfrac{17}{4}:\dfrac{23}{8}=\dfrac{17}{4}\cdot\dfrac{8}{23}=\dfrac{34}{23}\)
=>\(x=\dfrac{34}{23}\cdot1.61=2,38\)
c: -5x=2y
=>\(\dfrac{x}{2}=\dfrac{y}{-5}\)
mà x-y=-21
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-21}{7}=-3\)
=>\(\left\{{}\begin{matrix}x=-3\cdot2=-6\\y=\left(-3\right)\cdot\left(-5\right)=15\end{matrix}\right.\)
d: \(\dfrac{x}{2}=\dfrac{y}{3}\)
=>\(\dfrac{x}{8}=\dfrac{y}{12}\left(1\right)\)
\(\dfrac{y}{4}=\dfrac{z}{5}\)
=>\(\dfrac{y}{12}=\dfrac{z}{15}\left(2\right)\)
Từ (1),(2) suy ra \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
mà x+y-z=10
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)
=>\(\left\{{}\begin{matrix}x=2\cdot8=16\\y=2\cdot12=24\\z=2\cdot15=30\end{matrix}\right.\)
f: Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=k\)
=>x=2k; y=5k
xy=10
=>\(2k\cdot5k=10\)
=>\(10k^2=10\)
=>\(k^2=1\)
=>\(\left[{}\begin{matrix}k=1\\k=-1\end{matrix}\right.\)
TH1: k=1
=>\(\left\{{}\begin{matrix}x=2\cdot1=2\\y=5\cdot1=5\end{matrix}\right.\)
TH2: k=-1
=>\(\left\{{}\begin{matrix}x=2\cdot\left(-1\right)=-2\\y=5\cdot\left(-1\right)=-5\end{matrix}\right.\)