Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=k\Leftrightarrow a=2k;b=3k\)
\(ab=24\Leftrightarrow6k^2=24\Leftrightarrow k^2=2\\ \Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=4;b=6\\a=-4;b=-6\end{matrix}\right.\)
Ta có :
\(\dfrac{a}{2}=\dfrac{b}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\end{matrix}\right.\)
mà \(ab=24\)
\(\Rightarrow2k.3k=24\)
\(\Rightarrow6k^2=24\)
\(\Rightarrow k^2=2^2\)
\(\Rightarrow k=\left\{{}\begin{matrix}2\\-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=\dfrac{b}{3}=2\\\dfrac{a}{2}=\dfrac{b}{3}=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=4;b=6\\a=-4;b=-6\end{matrix}\right.\)
Ta có:
(a.b)/(2.3)= 24/6=4(T/c dãy tỉ số = nhau )
=>a=4.2=8
b=4.3=12
Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=k\) khi đó
\(\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\end{matrix}\right.\)
\(\Rightarrow a.b=2k.3k=6k^2\)
mà \(a.b=24\) ( bài cho)
\(\Rightarrow6k^2=24\)
\(\Rightarrow k^2=\dfrac{24}{6}\)
\(\Rightarrow k^2=4\)
\(\Rightarrow\left[{}\begin{matrix}k^2=2^2\\k^2=\left(-2\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
Th1: \(k=2\)
\(\Rightarrow\left\{{}\begin{matrix}a=2.2\\b=3.2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=4\\b=6\end{matrix}\right.\)
Th2: \(k=-2\)
\(\Rightarrow\left\{{}\begin{matrix}a=-2.2\\b=-2.3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=-4\\b=-6\end{matrix}\right.\)
Vậy \(\left(a;b\right)\in\left\{\left(4:6\right),\left(-4:-6\right)\right\}\)
