\(\frac{a}{b}=\frac{2}{5}\) nên a = 2k , b = 5k \((k\inℤ,k\ne0)\)
Từ a . b = 40 \(\Rightarrow2k\cdot5k=40\)
\(\Rightarrow k^2=40:(2\cdot5)\)
\(\Rightarrow k^2=40:10\)
\(\Rightarrow k^2=4\Rightarrow k=\pm2\)
Vậy : a = 4 b = 10 \((k=2)\)
a = -4 b = -10 \((k=-2)\)
Gọi hai số cần tìm a, b
Theo đề bài ta có
\(a:b=2:5\Leftrightarrow\frac{a}{2}=\frac{b}{5}\Leftrightarrow5a=2b\left(1\right)\)
\(a.b=40\Leftrightarrow a=\frac{40}{b}\left(2\right)\)
Thay (2) vào (1), ta có:
\(5.\frac{40}{b}=2b\)
\(\Leftrightarrow\frac{200}{b}=2b\)
\(\Leftrightarrow2b^2=200\)
\(\Leftrightarrow b^2=100\)
\(\Leftrightarrow b=\pm10\)
Nếu \(b=10\Rightarrow a=\frac{40}{b}=\frac{40}{10}=4\)
Nếu \(b=-10\Rightarrow a=\frac{40}{b}=\frac{40}{-10}=-4\)
Vậy \(a=4;b=10\)
\(a=-4;b=-10\)
ba=52 nên a = 2k , b = 5k (k\inℤ,k\ne0)(k∈Z,k̸=0)
Từ a . b = 40 \Rightarrow2k\cdot5k=40⇒2k⋅5k=40
\Rightarrow k^2=40:(2\cdot5)⇒k2=40:(2⋅5)
\Rightarrow k^2=40:10⇒k2=40:10
\Rightarrow k^2=4\Rightarrow k=\pm2⇒k2=4⇒k=±2
Vậy : a = 4 b = 10 (k=2)(k=2)
a = -4 b = -10 (k=-2)(k=−2)
