a: Ta có: 3x=5y
nên x/5=y/3
Đặt x/5=y/3=k
=>x=5k; y=3k
Ta có: xy=54
\(\Leftrightarrow15k^2=54\)
\(\Leftrightarrow k^2=3.6\)
Trường hợp 1: \(k=\dfrac{3\sqrt{10}}{5}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5k=\dfrac{15\sqrt{10}}{5}=3\sqrt{10}\\y=3k=\dfrac{9\sqrt{10}}{5}\end{matrix}\right.\)
Trường hợp 2: \(k=-\dfrac{3\sqrt{10}}{5}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5k=\dfrac{-15\sqrt{10}}{5}=-3\sqrt{10}\\y=3k=\dfrac{-9\sqrt{10}}{5}\end{matrix}\right.\)
b: 2x=3y
nên x/3=y/2
Đặt x/3=y/2=k
=>x=3k; y=2k
\(2x^3+y^3=62\)
\(\Leftrightarrow2\cdot27k^3+8k^3=62\)
=>k=1
=>x=3; y=2