Question

[Lê Mai]

1) \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2y=x^3+1\end{matrix}\right.\)

2) \(\left\{{}\begin{matrix}\sqrt{x^2+y^2}+\sqrt{2xy}=8\sqrt{2}\\\sqrt{x}+\sqrt{y}=4\end{matrix}\right.\)

3) \(\left\{{}\begin{matrix}\sqrt{x^3+3}+\left|y\right|=\sqrt{3}\\\sqrt{y^2+5}+\left|x\right|=\sqrt{x^2+5}\end{matrix}\right.\)

Answer 1

[Nguyễn Hoàng Minh]

\(1,HPT\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)+\left(\dfrac{1}{y}-\dfrac{1}{x}\right)=0\\2y=x^3+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\dfrac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\2y=x^3+1\end{matrix}\right.\\ \Leftrightarrow2y=y^3+1\Leftrightarrow y^3-2y+1=0\\ \Leftrightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{-1+\sqrt{5}}{2}\\y=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(0;0\right);\left(\dfrac{-1+\sqrt{5}}{2};\dfrac{-1+\sqrt{5}}{2}\right);\left(\dfrac{-1-\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right)\)

\(2,HPT\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(x^2+y^2\right)}+2\sqrt{xy}=16\\x+y+2\sqrt{xy}=16\end{matrix}\right.\\ \Leftrightarrow\sqrt{2\left(x^2+y^2\right)}=x+y\\ \Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x=y\\ \Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4\)

Vậy \(\left(x;y\right)=\left(4;4\right)\)

Answer 2

[Nguyễn Hoàng Minh]

\(3,\text{Sửa: }\left\{{}\begin{matrix}\sqrt{x^2+3}+\left|y\right|=\sqrt{3}\left(1\right)\\\sqrt{y^2+5}+\left|x\right|=\sqrt{x^2+5}\left(2\right)\end{matrix}\right.\)

Ta thấy \(\sqrt{x^2+3}\ge\sqrt{3};\left|y\right|\ge0\Leftrightarrow VT\left(1\right)\ge\sqrt{3}=VP\left(1\right)\)

Dấu \("="\Leftrightarrow x=y=0\)

Thay vào \(\left(2\right)\Leftrightarrow\sqrt{5}+0=\sqrt{5}\left(tm\right)\)

Vậy \(\left(x;y\right)=\left(0;0\right)\)