Question

\(Tìm\) \(x,y\) \(biết\) \(:\)

\(\frac{x}{2}=\frac{y}{5}\) \(và\) \(x.y=400\)

Answer 1

Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)

Thay \(x=2k;y=5k\) vào \(x.y=400\), ta có:

\(2k.5k=400\\ \Leftrightarrow10k^2=400\\ \Leftrightarrow k^2=40\\\Leftrightarrow k^2=\left(\pm\sqrt{40}\right)^2\\ \Rightarrow k\in\left\{\sqrt{40};-\sqrt{40}\right\}\)

+Khi \(k=\sqrt{40}\Rightarrow\left\{{}\begin{matrix}x=2.\sqrt{40}=4\sqrt{10}\\y=5.\sqrt{40}=10\sqrt{10}\end{matrix}\right.\)

+Khi \(k=-\sqrt{40}\left\{{}\begin{matrix}x=2.\left(-\sqrt{40}\right)=-4\sqrt{10}\\y=5.\left(-\sqrt{40}\right)=-10\sqrt{10}\end{matrix}\right.\)

Vậy...

Answer 2

\(\left\{{}\begin{matrix}\frac{x}{2}=\frac{y}{5}\\xy=400\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{2y}{5}\\xy=400\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=12,64911064\\y=31,6227766\end{matrix}\right.\)

Answer 3

Ta có:

\(\frac{x}{2}=\frac{y}{5}\) và \(x.y=400.\)

Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)

Có: \(x.y=400\)

=> \(2k.5k=400\)

=> \(10.k^2=400\)

=> \(k^2=400:10\)

=> \(k^2=40\)

=> \(k=?\)

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Answer 4

\(Tìm\) \(x\),\(y\) \(biết:\)

\(\frac{x}{2}\) \(=\) \(\frac{y}{5}\) và \(x\) . \(y\) \(=400\)

\(Bài\) \(làm\)

\(Đặt\) : \(u\) \(=\)\(\frac{x}{2}\)\(=\)\(\frac{y}{5}\)

\(Ta\) \(có\) :\(\frac{x}{2}\)\(=u\)⇒\(x=\)\(2u\)

\(\frac{y}{5}\) \(=u\) ⇒\(y=5u\)

\(Mà\) :\(x.y=400\)⇒ \(2u.5u=400\)⇒\(10u^2\)\(=400\)

⇒\(u^2\)\(=40\)

⇒\(u=20\) \(hoặc\) \(u=-20\)

\(Vậy\) \(ta\) \(có\) \(2\) \(trường\) \(hợp\):

\(TH1\) :\(u=20\) \(thì\) \(x=40\) ,\(y=100\)

\(TH2\) :\(u=-20\) \(thì\) \(x=-40\) ,\(y=-100\)

Answer 5

Sửa đề: \(\frac{x}{2}=\frac{y}{5}\) và \(x.y=40\)

Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)

Thay \(x=2k;y=5k\) vào \(x.y=40\) , ta có:

\(2k.5k=40\\ \Leftrightarrow10k^2=40\\ \Leftrightarrow k^2=4\\\Leftrightarrow k^2=\left(\pm2\right)^2\\ \Rightarrow k\in\left\{2;-2\right\} \)

+Khi \(k=2\Rightarrow\left\{{}\begin{matrix}x=2.2=4\\y=2.5=10\end{matrix}\right.\)

+Khi \(k=-2\Rightarrow\left\{{}\begin{matrix}x=-2.2=-4\\y=-2.5=-10\end{matrix}\right.\)

VẬy....