a: Đặt \(\dfrac{a}{5}=\dfrac{b}{4}=k\)
=>\(a=5k;b=4k\)
\(a^2-b^2=1\)
=>\(\left(5k\right)^2-\left(4k\right)^2=1\)
=>\(9k^2=1\)
=>\(k^2=\dfrac{1}{9}\)
=>\(\left[{}\begin{matrix}k=\dfrac{1}{3}\\k=-\dfrac{1}{3}\end{matrix}\right.\)
TH1: \(k=\dfrac{1}{3}\)
=>\(a=5k=\dfrac{5}{3};b=4k=\dfrac{4}{3}\)
TH2: \(k=-\dfrac{1}{3}\)
=>\(a=5k=5\cdot\dfrac{-1}{3}=-\dfrac{5}{3};b=4k=4\cdot\dfrac{-1}{3}=-\dfrac{4}{3}\)
b: Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\)
=>\(a=2k;b=3k;c=4k\)
\(a^2-b^2+2c^2=107-2ac\)
=>\(\left(2k\right)^2-\left(3k\right)^2+2\cdot\left(4k\right)^2+2\cdot2k\cdot4k=107\)
=>\(4k^2-9k^2+32k^2+16k^2=107\)
=>\(43k^2=107\)
=>\(k^2=\dfrac{107}{43}\)
=>\(\left[{}\begin{matrix}k=\sqrt{\dfrac{107}{43}}\\k=-\sqrt{\dfrac{107}{43}}\end{matrix}\right.\)
TH1: \(k=\sqrt{\dfrac{107}{43}}\)
=>\(a=2k=2\sqrt{\dfrac{107}{43}};b=3k=3\sqrt{\dfrac{107}{43}};c=4k=4\sqrt{\dfrac{107}{43}}\)
TH2: \(k=-\sqrt{\dfrac{107}{43}}\)
=>\(a=2k=-2\sqrt{\dfrac{107}{43}};b=3k=-3\sqrt{\dfrac{107}{43}};c=4k=-4\sqrt{\dfrac{107}{43}}\)