a) Đặt \(\frac{x}{3}=\frac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)
Thay \(x=3k,y=4k\) vào x . y = 84, có:
\(3k.4k=84 \\ \Leftrightarrow12k^2=84\\ \Leftrightarrow k^2=7\\ \Leftrightarrow k^2=\left(\pm\sqrt{7}\right)^2\\ \Rightarrow k\in\left\{\sqrt{7};-\sqrt{7}\right\}\)
+Khi \(k=\sqrt{7}\Rightarrow\left\{{}\begin{matrix}x=\sqrt{7}.3=3\sqrt{7}\\y=\sqrt{7}.4=4\sqrt{7}\end{matrix}\right.\)
+Khi \(k=-\sqrt{7}\Rightarrow\left\{{}\begin{matrix}x=-\sqrt{7}.3=-3\sqrt{7}\\y=-\sqrt{7}.4=-4\sqrt{7}\end{matrix}\right.\)
Vậy...
b)
Ta có: \(\frac{x}{5}=\frac{y}{4}.\)
\(\Rightarrow\frac{x^2}{25}=\frac{y^2}{16}\) và \(x^2-y^2=1.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{1}{9}.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x^2}{25}=\frac{1}{9}\Rightarrow x^2=\frac{25}{9}\Rightarrow\left[{}\begin{matrix}x=\frac{5}{3}\\x=-\frac{5}{3}\end{matrix}\right.\\\frac{y^2}{16}=\frac{1}{9}\Rightarrow y^2=\frac{16}{9}\Rightarrow\left[{}\begin{matrix}y=\frac{4}{3}\\y=-\frac{4}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\frac{5}{3};\frac{4}{3}\right),\left(-\frac{5}{3};-\frac{4}{3}\right).\)
Chúc bạn học tốt!