Câu 1:
a) Ta có: \(\frac{x}{2}=\frac{y}{3}\)
\(\Leftrightarrow\frac{4x}{8}=\frac{3y}{9}\)
Ta có: 4x-3y=-2
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{4x}{8}=\frac{3y}{9}=\frac{4x-3y}{8-9}=\frac{-2}{-1}=2\)
Do đó:
\(\left\{{}\begin{matrix}\frac{4x}{8}=2\\\frac{3y}{9}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x=16\\3y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)
Vậy: (x,y)=(4;6)
b) Đặt \(\frac{x}{4}=\frac{y}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)
Ta có: xy=20
\(\Leftrightarrow4k\cdot5k=20\)
\(\Leftrightarrow20k^2=20\)
\(\Leftrightarrow k^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}k=1\\k=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4k=4\cdot1=4\\y=5k=5\cdot1=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=4k=4\cdot\left(-1\right)=-4\\y=5k=5\cdot\left(-1\right)=-5\end{matrix}\right.\end{matrix}\right.\)
Vậy: (x,y)={(4;5);(-4;-5)}
Câu 2:
Ta có: \(\left\{{}\begin{matrix}\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\\x+y+z=9\end{matrix}\right.\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y+z}{2+3+4}=\frac{9}{9}=1\)
Do đó:
\(\left\{{}\begin{matrix}\frac{x}{2}=1\\\frac{y}{3}=1\\\frac{z}{4}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot1=2\\y=3\cdot1=3\\y=4\cdot1=4\end{matrix}\right.\)
Vậy: (x,y,z)=(2;3;4)
Câu 1
a) \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{4x}{8}=\frac{3y}{9}\)
\(\frac{4x}{8}=\frac{3y}{9}=\frac{4x-3y}{8-9}=\frac{-2}{-1}=2\)
\(\frac{4x}{8}=2\Rightarrow x=\frac{8.2}{4}=4\)
\(\frac{3y}{9}=2\Rightarrow y=\frac{2.9}{3}=6\)
Vậy: x = 4; y = 6
b) Đặt: \(\frac{x}{4}=\frac{y}{5}=k\)
Ta có:\(\left\{{}\begin{matrix}\frac{x}{4}=k\Rightarrow x=4k\\\frac{y}{5}=k\Rightarrow y=5k\end{matrix}\right.\)
\(x.y=20\)
=> 4k . 5k = 20
=> 20k = 20
=> k = 20 : 20 = 1
\(\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\)
Vậy: x = 4; y = 5
Câu 2:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y+z}{2+3+4}=\frac{9}{9}=1\)
Ta có: \(\left\{{}\begin{matrix}\frac{x}{2}=1\Rightarrow x=2.1=2\\\frac{y}{3}=1\Rightarrow y=3.1=3\\\frac{z}{4}=1\Rightarrow z=4.1=4\end{matrix}\right.\)
Vậy: x = 2; y = 3; z = 4