Với ab=90 => a=90/b
Thay a=90/b vào biểu thức \(\dfrac{a}{2}=\dfrac{b}{5}\) ta được :
\(\dfrac{\left(\dfrac{90}{b}\right)}{2}=\dfrac{b}{5}\) \(\Rightarrow\) \(\dfrac{90}{b}.5=b.2\Rightarrow\) \(\dfrac{450}{b}=2b\) \(\Rightarrow\) \(b^2=225\) \(\Rightarrow\) \(\left[{}\begin{matrix}b=15\\b=-15\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=6\\a=-6\end{matrix}\right.\)
Ta có :
\(ab=90\)
\(\dfrac{a}{2}=\dfrac{b}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2k\\b=5k\end{matrix}\right.\)
\(\Leftrightarrow a.b=2k.5k=10k^2=90\)
\(\Leftrightarrow k^2=9\)
\(\Leftrightarrow\left\{{}\begin{matrix}k^2=3^2\\k^2=\left(-3\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)
+) Với \(k=3\Leftrightarrow\left\{{}\begin{matrix}a=2.k=2.3=6\\b=5.k=5.3=15\end{matrix}\right.\)
+) Với \(k=-3\Leftrightarrow\left\{{}\begin{matrix}a=2.k=2.\left(-3\right)=-6\\b=5.k=5.\left(-3\right)=-15\end{matrix}\right.\)
Vậy ......
Giải.
Ta có : Từ \(\dfrac{a}{2}=\dfrac{b}{5}\Rightarrow\dfrac{a}{2}.b=\dfrac{b}{5}.b\)
\(\Rightarrow\dfrac{ab}{2}=\dfrac{b^2}{5}\)
lại có ab = 90\(\Rightarrow\dfrac{b^2}{5}=\dfrac{90}{2}=45\)
\(\Rightarrow b^2=225=\left(\pm15\right)^2\)
* Nếu \(b=15\Rightarrow a=6\)
* Nếu \(b=-15\Rightarrow a=-6\)
Vậy các cặp số (a,b) thỏa mãn là (6;15) và(-6;-15)
tik mik nha !!!
Đặt:
\(\dfrac{a}{2}=\dfrac{b}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=2k\\b=5k\end{matrix}\right.\)
\(\Rightarrow ab=2k.5k\)
\(\Rightarrow10k^2=90\Rightarrow k^2=9\Rightarrow k=\pm3\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=3.2=6\\b=3.5=15\end{matrix}\right.\\\left\{{}\begin{matrix}a=-3.2=-6\\b=-3.5=-15\end{matrix}\right.\end{matrix}\right.\)
