ta có \(\frac{x}{3}=\frac{y}{4}=\frac{z}{7}\)và x.y=48
xét \(\frac{x}{3}=\frac{y}{4}\)
đặt K vào \(\frac{x}{3}=\frac{y}{4}\)
ta có
\(\frac{x}{3}=K\Rightarrow x=3K\)
\(\frac{y}{4}=K\Rightarrow y=4K\)
\(x.y=48\)
\(3K.4K=48\)
\(12K^2=48\)
\(K^2=48:12=4\)
\(K^2=2^2\Rightarrow K=2\)
*\(\frac{x}{3}=2\Rightarrow x=2.3=6\)
*\(\frac{y}{4}=2\Rightarrow y=2.4=8\)
*\(\frac{z}{7}=2\Rightarrow z=2.7=14\)
vậy \(x=6;y=8;z=14\)
dat \(\frac{x}{3}=\frac{y}{4}=\frac{z}{7}=k\) => x=3k,y=4k,z=7k
Thay vvao ta dc: x.y=48
3k.4k=48
12.\(k^2\)=48
k^2=4
k=4,-4
TH1: k=a
=> x=3k=>x=12
y va z lam tuong tu nhe
Con TH2 la -4
k cho m nha
\(\frac{x}{3}=\frac{y}{4}=\frac{z}{7}\)Và \(x\cdot y=48\)
Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{7}=K\)
\(\Rightarrow\frac{x}{3}=K\Rightarrow x=3K\)
\(\Rightarrow\frac{y}{4}=K\Rightarrow y=4K\)
\(\Rightarrow\frac{z}{7}=K\Rightarrow z=7K\)
Mà \(x\cdot y=48\)
\(\Rightarrow3K\cdot4k=48\)
\(\Rightarrow12K^2=48\)
\(\Rightarrow K^2=4\)
\(\Rightarrow K=2\)
Khi đó: \(\Rightarrow\frac{x}{3}=2\Rightarrow x=6\)
\(\Rightarrow\frac{y}{4}=2\Rightarrow y=8\)
\(\Rightarrow\frac{z}{7}=2\Rightarrow z=14\)
Vậy x=3;y=8 và z=14
