1. Áp dụng tc dãy TSBN, ta có:
\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y-z}{6+5-3}=\dfrac{54}{8}=\dfrac{27}{4}\)
+\(\dfrac{x}{6}=\dfrac{27}{4}\Rightarrow x=\dfrac{27.6}{4}=\dfrac{81}{2}\)
+\(\dfrac{y}{5}=\dfrac{27}{4}\Rightarrow y=\dfrac{27.5}{4}=\dfrac{135}{4}\)
+\(\dfrac{z}{3}=\dfrac{27}{4}\Rightarrow z=\dfrac{27.3}{4}=\dfrac{81}{4}\)
Vậy \(x=\dfrac{81}{2};y=\dfrac{135}{4};z=\dfrac{81}{4}\)
2,Áp dụng tc dãy TSBN, ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{c}{4}=\dfrac{x+2y-3c}{2+2.3+3.4}=\dfrac{-20}{20}=-1\)
+\(\dfrac{x}{2}=-1\Rightarrow x=-1.2=-2\)
+\(\dfrac{y}{3}=-1\Rightarrow y=-1.3=-3\)
+\(\dfrac{c}{4}=-1\Rightarrow c=-1.4=-4\)
Vậy \(x=-2;y=-3;c=-4\)
3,Từ \(5x=8y=20z\Rightarrow\dfrac{5x}{160}=\dfrac{8y}{160}=\dfrac{20z}{160}\)
\(\Rightarrow\dfrac{x}{32}=\dfrac{y}{20}=\dfrac{z}{8}\)
Áp dụng tc dãy TSBN, ta có:
\(\dfrac{x}{32}=\dfrac{y}{20}=\dfrac{z}{8}=\dfrac{x-y-z}{32-20-8}=\dfrac{3}{4}\)
+\(\dfrac{x}{32}=\dfrac{3}{4}\Rightarrow x=\dfrac{32.3}{4}=24\)
+\(\dfrac{y}{20}=\dfrac{3}{4}\Rightarrow y=\dfrac{20.3}{4}=15\)
+\(\dfrac{z}{8}=\dfrac{3}{4}\Rightarrow z=\dfrac{3.8}{4}=6\)
Vậy \(x=24;y=15;z=6\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{6}=\frac{y}{5}=\frac{z}{3}=\frac{x+y-z}{6+5-3}=\frac{54}{8}=\frac{27}{4}\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\) (k≠0)
⇒x=3k
y=4k
Ta có:xy=48
⇒3k.4k=48
12.\(k^2=48\)
\(k^2=48:12=4\)
⇒ k∈{4;-4]
TH1:k=4
\(\Rightarrow\left[{}\begin{matrix}x=3k=3.4=12\\y=4k=4.4=16\end{matrix}\right.\)
TH2:k=-4
\(\Rightarrow\left[{}\begin{matrix}x=3k=3.\left(-4\right)=-12\\y=4k=4.\left(-4\right)=-16\end{matrix}\right.\)
Vậy x=12 thì y=16 ; x=-12 thì y=-17