a, \(\dfrac{x}{y}=\dfrac{17}{3}\&x+y=-60\)
Từ \(\dfrac{x}{y}=\dfrac{17}{3}\Rightarrow\dfrac{x}{17}=\dfrac{y}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x}{17}=\dfrac{y}{3}=\dfrac{x+y}{17+3}=\dfrac{-60}{20}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{17}=-3\\\dfrac{y}{3}=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-51\\y=-9\end{matrix}\right.\)
b, \(\dfrac{x}{19}=\dfrac{y}{21}\&2x-y=34\)
\(\dfrac{x}{19}=\dfrac{y}{21}\Leftrightarrow\dfrac{2x}{2.19}=\dfrac{y}{21}\Leftrightarrow\dfrac{2x}{38}=\dfrac{y}{21}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x}{38}=2\\\dfrac{y}{21}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=38\\y=42\end{matrix}\right.\)
c, \(\dfrac{x^2}{9}=\dfrac{y^2}{16}\&x^2+y^2=100\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{100}{25}=4\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{9}=4\\\dfrac{y^2}{16}=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=54\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\\\left[{}\begin{matrix}y=8\\y=-8\end{matrix}\right.\end{matrix}\right.\)
Các cặp (x;y) tương ứng là: \(\left(6;8\right)\&\left(-6;-8\right)\)
a) x/y=17/3=>x/17=y/3=x+y/17+3=60/20=3
Vậy: x=3.17=51; y=3.3=9
b)x/19=y/21=2x-y/19.2-21=34/17=2
Vậy: x=2.19=38; y=2.21=42
c)x^2/9=y^2/16=x^2+y^2/9+16=100/25=4
Vậy: x^2=36=>x=6; y^2=64=>y=8
