a/ Đặt :
\(\frac{x}{12}=\frac{y}{9}=k\Leftrightarrow\left\{{}\begin{matrix}x=12k\\y=9k\end{matrix}\right.\)
Mà \(xy=12\)
\(\Leftrightarrow12k.9k=12\)
\(\Leftrightarrow k^2=\frac{1}{9}\) \(\Leftrightarrow\left[{}\begin{matrix}k=\frac{1}{3}\\k=-\frac{1}{3}\end{matrix}\right.\)
+) \(k=\frac{1}{3}\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\)
+) \(k=-\frac{1}{3}\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=-3\end{matrix}\right.\)
Vậy ..
b/ Ta có :
\(\frac{x}{y}=\frac{9}{7}\Leftrightarrow\frac{x}{9}=\frac{y}{7}\left(1\right)\)
\(3y=7z\Leftrightarrow\frac{y}{7}=\frac{z}{3}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\frac{x}{9}=\frac{y}{7}=\frac{z}{3}\)
Theo t/c dãy tỉ số bằng nhau ta có :
\(\frac{x}{9}=\frac{y}{7}=\frac{z}{3}=\frac{x-y+z}{9-7+3}=\frac{-15}{5}=-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{9}=-3\\\frac{y}{7}=-3\\\frac{z}{3}=-3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)
Vậy...
a, x/12=y/9 và x.y =12
a/12=y/9=a.y/12.9=12/108=1/9
a/12=1/9=>a=1/9.12=4/3
y/9=1/9=>y=1/9.9=1
câu b mik bó tay
a) Ta có: \(\frac{x}{12}=\frac{y}{9}.\)
=> \(\frac{x}{12}=\frac{y}{9}\) và \(x.y=12.\)
Đặt \(\frac{x}{12}=\frac{y}{9}=k\Rightarrow\left\{{}\begin{matrix}x=12k\\y=9k\end{matrix}\right.\)
Có: \(x.y=12\)
=> \(12k.9k=12\)
=> \(108k^2=12\)
=> \(k^2=12:108\)
=> \(k^2=\frac{1}{9}\)
=> \(k=\pm\frac{1}{3}.\)
TH1: \(k=\frac{1}{3}.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{3}.12=4\\y=\frac{1}{3}.9=3\end{matrix}\right.\)
TH2: \(k=-\frac{1}{3}.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-\frac{1}{3}\right).12=-4\\y=\left(-\frac{1}{3}\right).9=-3\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(4;3\right),\left(-4;-3\right).\)
Chúc bạn học tốt!
