2x=3y=4z =k
suy ra x=k/2; y=k/3, z=k/4
mà xy + yz + zx = 6
suy ra \(\frac{k^2}{6}+\frac{k^2}{12}+\frac{k^2}{8}=6\Rightarrow k^2.\frac{3}{8}=6\Rightarrow k^2=16\Rightarrow k\in\left\{4;-4\right\}\)
Với k = 4 suy ra x =2; y=4/3; z=1
Với k =- 4 suy ra x =-2; y=-4/3; z=-1
Ta có :
\(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\)
\(\Leftrightarrow\frac{x}{6}=\frac{y}{4}\)
\(3y=4z\Leftrightarrow\frac{z}{3}=\frac{y}{4}\)
\(\Rightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\)
Ta có :
\(\left(\frac{x}{6}\right)^2=\frac{x}{6}.\frac{x}{6}=\frac{x}{6}.\frac{y}{4}=\frac{y}{4}.\frac{z}{3}=\frac{z}{3}.\frac{y}{6}\)
\(\Leftrightarrow\)\(\left(\frac{x}{6}\right)^2\)\(=\frac{xy}{24}=\frac{yz}{12}=\frac{zx}{18}=\frac{xy+yz+zx}{24+12+18}=\frac{1}{9}\)\(\left(\text{T/c dãy tỉ số bằng nhau}\right)\)
\(\Rightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\)\(=\pm\frac{1}{3}\)
\(Th1:\hept{\begin{cases}x=2\\y=\frac{4}{3}\\z=1\end{cases}}\)
\(Th2:\hept{\begin{cases}x=-2\\y=-\frac{4}{3}\\z=-1\end{cases}}\)
Trả lời:
\(2x=3y=4z\)
\(\Leftrightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\)
Đặt\(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}=k\)
\(\Rightarrow\hept{\begin{cases}x=6k\\y=4k\\z=3k\end{cases}}\)
Mà\(xy+yz+zx=6\)
\(\Rightarrow6k.4k+4k.3k+3k.6k=6\)
\(\Leftrightarrow24k^2+12k^2+18k^2=6\)
\(\Leftrightarrow54k^2=6\)
\(\Leftrightarrow k^2=\frac{1}{9}\)
\(\Leftrightarrow\orbr{\begin{cases}k=\frac{1}{3}\\k=\frac{-1}{3}\end{cases}}\)
+ Với\(k=\frac{1}{3}\)
\(\Rightarrow\hept{\begin{cases}x=2\\y=\frac{4}{3}\\z=1\end{cases}}\)(Thỏa mãn)
+ Với\(k=\frac{-1}{3}\)
\(\Rightarrow\hept{\begin{cases}x=-2\\y=\frac{-4}{3}\\z=-1\end{cases}}\)(Thỏa mãn)
Vậy\(\hept{\begin{cases}x=2\\y=\frac{4}{3}\\z=1\end{cases}}\)hoặc\(\hept{\begin{cases}x=-2\\y=\frac{-4}{3}\\z=-1\end{cases}}\)
Hok tốt!
Good girl
