Đặt a/3=k=>a=3k
b/4=k=>b=4k
Ta có: a.b=48
<=> 3k.4k=48
<=> 12k^2=48
<=> k^2=4
<=> k=2 hoặc k=-2
Với k=2=>a=3.2=6; b=4.2=8
Với k=-2=>a=3.(-2)=-6; b=4.(-2)=-8
Vậy a=6 hoặc -6
b=8 hoặc -8
Cách #:
Ta có: \(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{ab}{3b}=\dfrac{48}{3b}\)
\(\Rightarrow\dfrac{b}{4}=\dfrac{48}{3b}\Rightarrow b\cdot3b=48\cdot4\Rightarrow3b^2=192\)
\(\Rightarrow b^2=\dfrac{192}{3}=64\Rightarrow\left[{}\begin{matrix}b=8\\b=-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{48}{8}=6\\a=\dfrac{48}{-8}=-6\end{matrix}\right.\)
Vậy \(\left(a;b\right)=\left(6;8\right);\left(-6;-8\right)\)
Ta có (a/3)2=a/3 x a/3=a/3 x b/4 =a.b/3.4=48/12=4
=> a=12; b=16
