a) Ta có :
\(2y-3y+z=6\)
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x}{9}=\dfrac{y}{12}\)
\(\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{y}{12}=\dfrac{y}{20}\)
\(\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=3\Leftrightarrow x=27\\\dfrac{y}{12}=3\Leftrightarrow y=36\\\dfrac{z}{20}=3\Leftrightarrow z=60\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=27\\y=36\\z=60\end{matrix}\right.\) là giá trị cần tìm
Đặt :
\(\dfrac{x}{2}=\dfrac{y}{5}=k\) \(\left(k\in N\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)
Ta có :
\(xy=40\)
\(\Leftrightarrow2k.5k=40\)
\(\Leftrightarrow10.k^2=40\)
\(\Leftrightarrow k^2=4\)
\(\Leftrightarrow\left\{{}\begin{matrix}k^2=2^2\\k^2=\left(-2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
+) \(k=2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2.2=4\\y=5k=5.2=10\end{matrix}\right.\)
+) \(k=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2.\left(-2\right)=-4\\y=5k=5.\left(-2\right)=-10\end{matrix}\right.\)
Vậy \(\left(x;y\right)\) cần tìm là \(\left(4;10\right),\left(-4;-10\right)\)
Câu a :
Ta có :
\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{3}=\dfrac{x}{5}\) \(\Leftrightarrow\) \(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\Leftrightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\)
Áp dụng t/c dãy tỉ số = nhau ta có :
\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{9}=3\Rightarrow x=27\\\dfrac{y}{12}=3\Rightarrow y=36\\\dfrac{z}{20}=3\Rightarrow z=60\end{matrix}\right.\)
Vậy ......................
Câu b : Câu này thấy nó ngộ ngộ :D
\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{3}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{20}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
\(\Rightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\)
ÁP dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\)
\(=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.9=27\\y=3.12=36\\z=3.20=60\end{matrix}\right.\)
b) Đặt: \(\dfrac{x}{2}=\dfrac{y}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)
\(\Rightarrow2k.5k=40\)
\(\Rightarrow10k^2=40\)
\(\Rightarrow k^2=4\Rightarrow k=\pm2\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2.2=4\\y=2.5=10\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2.2=-4\\y=-2.5=-10\end{matrix}\right.\end{matrix}\right.\)
