Question

\(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=16\\2\sqrt{x}-3\sqrt{y}=-11\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=10\\2\sqrt{x}-3\sqrt{y}=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6\sqrt{x}+4\sqrt{y}=20\\6\sqrt{x}-9\sqrt{y}=-33\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}13\sqrt{y}=53\\3\sqrt{x}+2\sqrt{y}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2809}{169}\\x=\dfrac{64}{169}\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(\dfrac{64}{169};\dfrac{2809}{169}\right)\)

Answer 2

\(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=16\\2\sqrt{x}-3\sqrt{y}=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6\sqrt{x}+4\sqrt{y}=32\\6\sqrt{x}-9\sqrt{y}=-33\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}13\sqrt{y}=65\\3\sqrt{x}+2\sqrt{y}=16\end{matrix}\right.

\(\Leftrightarrow\left\{{}\begin{matrix}y=25\\x\notin\varnothing\end{matrix}\right.\)

Vậy nghiệm hệ phương trình vô nghiệm

Answer 3

\(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=16\\2\sqrt{x}-3\sqrt{y}=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6\sqrt{x}+4\sqrt{y}=32\\6\sqrt{x}-9\sqrt{y}=-33\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}13\sqrt{y}=65\\3\sqrt{x}+2\sqrt{y}=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=25\\x\(\in\varnothing\)\end{matrix}\right.\)

Vậy hệ phương trình vô nghiệm

Answer 4

\(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=16\\2\sqrt{x}-3\sqrt{y}=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6\sqrt{x}+4\sqrt{y}=32\\6\sqrt{x}-9\sqrt{y}=-33\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}13\sqrt{y}=65\\3\sqrt{x}+2\sqrt{y}=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=25\\x\in\varnothing\end{matrix}\right.\)

Vậy hệ phương trình vô nghiệm

Answer 5

\(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=16\\2\sqrt{x}-3\sqrt{y}=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6\sqrt{x}+4\sqrt{y}=32\left(1\right)\\6\sqrt{x}-9\sqrt{y}=-33\left(2\right)\end{matrix}\right.\)

Cộng (1) và (2), ta được phương trình:

\(13\sqrt{y}=65\Rightarrow y=25\)

Thay $y=25$ vào (1) ta được:

\(6\sqrt{x}+4\sqrt{25}=32\\ \Rightarrow6\sqrt{x}+20=32\\ \Rightarrow6\sqrt{x}=12\\ \Rightarrow x=4\)

Vậy nghiệm hệ phương trình là (4; 25)