\(a,3x=5y\)và \(xy=60\)
\(3x=5y\)
\(\Leftrightarrow\frac{x}{5}=\frac{y}{3}\)
\(\Leftrightarrow\frac{x^2}{25}=\frac{xy}{15}=\frac{y^2}{9}\)
\(\Leftrightarrow\frac{x^2}{25}=\frac{y^2}{9}=\frac{60}{15}=4\)
\(\Rightarrow\hept{\begin{cases}\frac{x^2}{25}=4\\\frac{y^2}{9}=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2=100\\y^2=36\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\pm10\\y=\pm6\end{cases}}\)
Vậy \(\left(x,y\right)\in\left\{\left(-10,-6\right);\left(10,6\right)\right\}\)
\(b,4x=5y\)và \(x^2-y^2=9\)
\(4x=5y\)
\(\Leftrightarrow\frac{x}{5}=\frac{y}{4}\)
\(\Leftrightarrow\frac{x^2}{25}=\frac{y^2}{16}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{9}{9}=1\)
\(\Rightarrow\hept{\begin{cases}\frac{x^2}{25}=1\\\frac{y^2}{16}=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2=25\\y^2=16\end{cases}\Leftrightarrow\hept{\begin{cases}x=\pm5\\y=\pm4\end{cases}}}\)
Vậy \(\left(x,y\right)\in\left\{\left(-5,-4\right);\left(5,4\right)\right\}\)
\(c,x:3=y:7\)và xy = 21
\(x:3=y:7\)
\(\Leftrightarrow\frac{x}{3}=\frac{y}{7}\)
\(\Leftrightarrow\frac{x^2}{9}=\frac{xy}{21}=\frac{y^2}{49}\)
\(\Leftrightarrow\frac{x^2}{9}=\frac{y^2}{49}=\frac{21}{21}=1\)
\(\Rightarrow\hept{\begin{cases}\frac{x^2}{9}=1\\\frac{y^2}{49}=1\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=9\\y^2=49\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\pm3\\y=\pm7\end{cases}}}\)
Vậy \(\left(x,y\right)\in\left\{\left(-3,-7\right);\left(3,7\right)\right\}\)
\(d,2x=9y\)và xy = 72
\(2x=9y\)
\(\Leftrightarrow\frac{x}{9}=\frac{y}{2}\)
\(\Leftrightarrow\frac{x^2}{81}=\frac{xy}{18}=\frac{y^2}{4}\)
\(\Leftrightarrow\frac{x^2}{81}=\frac{y^2}{4}=\frac{72}{18}=4\)
\(\Rightarrow\hept{\begin{cases}\frac{x^2}{81}=4\\\frac{y^2}{4}=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2=324\\y^2=16\end{cases}\Leftrightarrow\hept{\begin{cases}x=\pm18\\y=\pm4\end{cases}}}\)
Vậy \(\left(x,y\right)\in\left\{\left(-18,-4\right);\left(18,4\right)\right\}\)
\(a;c;d\) AD vs k
\(b,4x=5y\)và \(x^2-y^2=9\)
Theo bài ra ta cs
\(4x=5y\Rightarrow\frac{x}{5}=\frac{y}{4}\)và \(x^2-y^2=9\)
ADTC dãy tỉ số bằng nhau
\(\frac{x}{5}=\frac{y}{4}=\frac{x^2-y^2}{5^2-4^2}=\frac{9}{9}=1\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=1\\\frac{y}{4}=1\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x=4\end{cases}}}\)