P) $\left\{\begin{matrix} 4\sqrt{5}x-y=3\sqrt{2}\\\sqrt{10}x+\sqrt{2}y=-1 \end{matrix}\right.$

`<=>` $\left\{\begin{matrix} 4\sqrt{10}x-\sqrt{2}y=6\\\sqrt{10}x+\sqrt{2}y=-1 \end{matrix}\right.$

`<=>` $\left\{\begin{matrix} 5\sqrt{10}x=5\\\sqrt{10}x+\sqrt{2}y=-1 \end{matrix}\right.$

`<=>` $\left\{\begin{matrix} \sqrt{10}x=1\\\sqrt{10}x+\sqrt{2}y=-1 \end{matrix}\right.$

`<=>` $\left\{\begin{matrix} x=\dfrac{1}{\sqrt{10}}\\\sqrt{10}x+\sqrt{2}y=-1 \end{matrix}\right.$

`<=>` $\left\{\begin{matrix} x=\dfrac{\sqrt{10}}{10}\\\sqrt{10}·\dfrac{\sqrt{10}}{10}+\sqrt{2}y=-1 \end{matrix}\right.$

`<=>` $\left\{\begin{matrix} x=\dfrac{\sqrt{10}}{10}\\\sqrt{2}y=-2 \end{matrix}\right.$

`<=>` $\left\{\begin{matrix} x=\dfrac{\sqrt{10}}{10}\\y=-\sqrt{2} \end{matrix}\right.$

Vậy hệ `pt` có nghiệm  `(x;y)=(\frac{\sqrt{10}}{10};-\sqrt{2})`

Q) $\left\{\begin{matrix} \sqrt{2}x-\sqrt{3}y=1\\5\sqrt{2}x-4\sqrt{3}y=8 \end{matrix}\right.$

`<=>` $\left\{\begin{matrix} 4\sqrt{2}x-4\sqrt{3}y=4\\5\sqrt{2}x-4\sqrt{3}y=8 \end{matrix}\right.$

`<=>` $\left\{\begin{matrix} -\sqrt{2}x=-4\\\sqrt{2}x-\sqrt{3}y=1\end{matrix}\right.$

`<=>` $\left\{\begin{matrix} x=2\sqrt{2}\\\sqrt{2}·2\sqrt{2}-\sqrt{3}y=1\end{matrix}\right.$

`<=>` $\left\{\begin{matrix} x=2\sqrt{2}\\-\sqrt{3}y=1-4\end{matrix}\right.$

`<=>` $\left\{\begin{matrix} x=2\sqrt{2}\\-\sqrt{3}y=-3\end{matrix}\right.$

`<=>` $\left\{\begin{matrix} x=2\sqrt{2}\\y=\sqrt{3}\end{matrix}\right.$

Vậy hệ `pt` có nghiệm  `(x;y)=(2\sqrt{2};\sqrt{3})`

Đáp án + giải thích các bước giải:

p) $\left\{\begin{matrix} 4\sqrt{5}x-y=3\sqrt{2}\\\sqrt{10}x+\sqrt{2}y=-1 \end{matrix}\right.\\\to \left\{\begin{matrix} 4\sqrt{10}x-\sqrt{2}y=6\\\sqrt{10}x+\sqrt{2}y=-1 \end{matrix}\right.\\\to \left\{\begin{matrix} 4\sqrt{10}x-\sqrt{2}y+\sqrt{10}x+\sqrt{2}y=6-1\\\sqrt{10}x+\sqrt{2}y=-1 \end{matrix}\right. \\\to \left\{\begin{matrix} 5\sqrt{10}x=5 \\ \sqrt{10}x+\sqrt{2}y=-1 \end{matrix}\right. \\\to \left\{\begin{matrix} x=\dfrac{1}{\sqrt{10}}\\ \sqrt{10}x+\sqrt{2}y=-1 \end{matrix}\right. \\\to \left\{\begin{matrix} x=\dfrac{\sqrt{10}}{10}\\\sqrt{2}y=-2\end{matrix}\right. \\\to \left\{\begin{matrix} x=\dfrac{\sqrt{10}}{10}\\y=-\sqrt{2}\end{matrix}\right.$

Vậy $(x;y)=\bigg(\dfrac{\sqrt{10}}{10};-\sqrt{2}\bigg)$

q) $\left\{\begin{matrix} \sqrt{2}x-\sqrt{3}y=1\\5\sqrt{2}x-4\sqrt{3}=8\end{matrix}\right. \\\to \left\{\begin{matrix} 5\sqrt{2}x-5\sqrt{3}y=5\\5\sqrt{2}x-4\sqrt{3}=8\end{matrix}\right.\\\to \left\{\begin{matrix} 5\sqrt{2}x-5\sqrt{3}y-(5\sqrt{2}x-4\sqrt{3}y)=5-8\\\sqrt{2}x-\sqrt{3}y=1\end{matrix}\right. \\\to \left\{\begin{matrix} -\sqrt{3}y=-3\\\sqrt{2}x-\sqrt{3}y=1\end{matrix}\right. \\\to \left\{\begin{matrix} y=\sqrt{3}\\\sqrt{2}x=4 \end{matrix}\right. \\\to \left\{\begin{matrix} y=\sqrt{3}\\x=2\sqrt{2}\end{matrix}\right.$

Vậy $(x;y)=(\sqrt{3};2\sqrt{2})$