Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=2.k\\b=3.k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a.b=150.k^2\\doa.b=150\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}150.k^2=150\\k^2=150:150\\k^2=1\\\rightarrow k=\pm1\end{matrix}\right.\)
Với: k=1 => \(\left\{{}\begin{matrix}a=2.1=2\\b=3.1=3\end{matrix}\right.\)
Với : k = (-1) => \(\left\{{}\begin{matrix}a=2.\left(-1\right)=-2\\b=3.\left(-1\right)=-3\end{matrix}\right.\)
Kết luận : ( a,b) = ( 2;3 ) , ( -2 ; -3 )
\(\)Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=k\)
=> \(a=2k,b=3k\)
Ta có : \(a.b=2k.3k=150\)
=> \(a.b=6.k^2=150\)
=> \(k^2=150:6\)
=> \(k^2=25=>k=5\) hay \(k=-5\)
Với \(k=5\) => \(a=2k=2.5=10\) ; \(b=3k=3.5=15\)
Với \(k=-5\)=> \(a=2k=2.\left(-5\right)=-10\) ; \(b=3k=3.\left(-5\right)=-15\)
Vậy a = 10, b = 15 hay a = - 10 , b = - 15
