\(P=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}=\sqrt{2}\left(\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{6}\right)\)
\(=\sqrt{2}\left(\frac{3+\sqrt{3}+3-\sqrt{3}}{6}\right)=\sqrt{2}\)
2/ ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{y^2-\frac{7}{y^2}}=a\\\sqrt{y-\frac{7}{y^2}}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=y\\a^2-b^2=y^2-y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=y\\\left(a-b\right)\left(a+b\right)=y^2-y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=y\\a-b=y-1\end{matrix}\right.\)
\(\Rightarrow b=\frac{1}{2}\Rightarrow\sqrt{y-\frac{7}{y^2}}=\frac{1}{2}\Rightarrow y-\frac{7}{y^2}=\frac{1}{4}\Rightarrow4y^3-y^2-28=0\)
\(\Rightarrow y=2\)
3/ \(\Leftrightarrow\left\{{}\begin{matrix}4x^2-2y^2=2\\xy+x^2=2\end{matrix}\right.\)
\(\Rightarrow3x^2-xy-2y^2=0\)
\(\Rightarrow\left(x-y\right)\left(3x+2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\x=-\frac{2}{3}y\end{matrix}\right.\) thay vào 1 trong 2 pt là xong
