\(\dfrac{a}{5}=\dfrac{b}{4}\Rightarrow\dfrac{a^2}{25}=\dfrac{b^2}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a^2}{25}=\dfrac{b^2}{16}=\dfrac{a^2-b^2}{25-16}=\dfrac{1}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^2}{25}=\dfrac{1}{9}\\\dfrac{b^2}{16}=\dfrac{1}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2=\dfrac{25}{9}\\b^2=\dfrac{16}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5}{3}\\b=\dfrac{4}{3}\end{matrix}\right.\)
Đặt : \(\dfrac{a}{5}=\dfrac{b}{4}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5k\\b=4k\end{matrix}\right.\)
Mà \(a^2-b^2=1\)
\(\Leftrightarrow\left(5k\right)^2-\left(4k\right)^2=1\)
\(\Leftrightarrow25k^2-16k^2=1\)
\(\Leftrightarrow9k^2=1\)
\(\Leftrightarrow k^2=\dfrac{1}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}k=\dfrac{1}{9}\\k=-\dfrac{1}{9}\end{matrix}\right.\)
+) \(k=\dfrac{1}{9}\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5}{9}\\b=\dfrac{4}{9}\end{matrix}\right.\)
+) \(k=-\dfrac{1}{9}\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{5}{9}\\b=-\dfrac{4}{9}\end{matrix}\right.\)
Vậy..
