Question

tìm các số x,y,z biết rằng \(\frac{x}{3}\)=\(\frac{y}{4}\);\(\frac{y}{5}\)=\(\frac{z}{6}\) và x.y =300

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Answer 1

Ta có:

\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}.\)

\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}.\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\) và \(x.y=300.\)

Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\Rightarrow\left\{{}\begin{matrix}x=15k\\y=20k\\z=24k\end{matrix}\right.\)

Lại có: \(x.y=300\)

\(\Rightarrow15k.20k=300\)

\(\Rightarrow300.k^2=300\)

\(\Rightarrow k^2=300:300\)

\(\Rightarrow k^2=1\)

\(\Rightarrow k^2=\left(\pm1\right)^2\)

\(\Rightarrow k=\pm1.\)

+ TH1: \(k=1.\)

\(\Rightarrow\left\{{}\begin{matrix}x=15.1=15\\y=20.1=20\\z=24.1=24\end{matrix}\right.\)

+ TH2: \(k=-1.\)

\(\Rightarrow\left\{{}\begin{matrix}x=15.\left(-1\right)=-15\\y=20.\left(-1\right)=-20\\z=24.\left(-1\right)=-24\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(15;20;24\right),\left(-15;-20;-24\right).\)

Chúc bạn học tốt!

Answer 2

Vì BCNN (4; 5) = 20

\(\Rightarrow\frac{x}{3}=\frac{y}{4}\Leftrightarrow\frac{x}{3.5}=\frac{y}{4.5}=\frac{x}{15}=\frac{y}{20}\)

\(\Rightarrow\frac{y}{5}=\frac{z}{6}\Leftrightarrow\frac{y}{5.4}=\frac{z}{6.4}=\frac{y}{20}=\frac{z}{24}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)

Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\Rightarrow\left\{{}\begin{matrix}x=15k\\y=20k\\z=24k\end{matrix}\right.\)

Thay \(x=15k,y=20k\) vào x . y = 300, có:

\(15k.20k=300\\ \Leftrightarrow300k^2=300\\ \Leftrightarrow k^2=1\\ \Rightarrow k\in\left\{1;-1\right\}\)

+ Khi \(k=1\Rightarrow\left\{{}\begin{matrix}x=1.15=15\\y=1.20=20\\z=1.24=24\end{matrix}\right.\)

+ Khi \(k=-1\Rightarrow\left\{{}\begin{matrix}x=-1.15=-15\\y=-1.20=-20\\z=-1.24=-24\end{matrix}\right.\)

Vậy...