Question

Tìm x và y, biết: \(x+y+13=2\left(2\sqrt{x}+3\sqrt{y}\right)\)

Answer 1

\(x+y+13=2\left(2\sqrt{x}+3\sqrt{y}\right)\)

\(\Leftrightarrow x+y+13-4\sqrt{x}-6\sqrt{y}=0\)

\(\Leftrightarrow x-4\sqrt{x}+4+y-6\sqrt{y}+9=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y}-3\right)^2=0\)

Xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x}=2\\\sqrt{y}=3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=9\end{matrix}\right.\)

Answer 2

Điều kiện xác định: \(x,y>0\)

\(x+y+13=2\left(2\sqrt{x}+3\sqrt{y}\right)\)

\(\Leftrightarrow x-4\sqrt{x}+4+y-6\sqrt{y}+9=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y}-3\right)^2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)^2=0\\\left(\sqrt{y}-3\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{\sqrt{x}-2=0}{\sqrt{y}-3=0}\)

\(\Leftrightarrow\dfrac{x=4}{y=9}\)

KL: x = 4, y = 9

Answer 3

Viết mở bài của câu chuyện về cuộc giao chiến của Sơn Tinh và Thuỷ Tinh

Answer 4

Cx khá dễ:

\(x+y+13=2\left(2\sqrt{x}+3\sqrt{y}\right)\)

\(\Rightarrow x+y+13=4\sqrt{x}+6\sqrt{y}\)

\(\Rightarrow x+y+13-4\sqrt{x}-6\sqrt{y}=0\)

\(\Rightarrow x+y+4+9-4\sqrt{x}-6\sqrt{y}=0\)

\(\Rightarrow\left(x-4\sqrt{x}+4\right)+\left(y-6\sqrt{y}+9\right)=0\)

\(\Rightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y}-3\right)^2=0\)

\(\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)^2\ge0\\\left(\sqrt{y}-3\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y}-3\right)^2\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\sqrt{x}=2\Rightarrow x=4\\\sqrt{y}=3\Rightarrow y=9\end{matrix}\right.\)

Answer 5

\(\Leftrightarrow\)x+y+13-2(2\(\sqrt{x}\)+3\(\sqrt{y}\))=0

\(\Leftrightarrow\)x+y+13-4\(\sqrt{x}\)-6\(\sqrt{y}\)=0

\(\Leftrightarrow\)x+y+4+9-4\(\sqrt{x}\)-6\(\sqrt{y}\)=0

\(\Leftrightarrow\)(\(\sqrt{x}\)-2)²+(\(\sqrt{y}\)-3)²

^{\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)^2=0\\\left(\sqrt{y}-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{y}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\sqrt{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=9\end{matrix}\right.\)}