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\(\Rightarrow\left\{{}\begin{matrix}x+y=\sqrt{2}\\xy=\frac{1}{4}\end{matrix}\right.\)
\(x^2+y^2=\left(x+y\right)^2-2xy=\frac{3}{2}\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=2\sqrt{2}-\frac{3}{4}.\sqrt{2}=\frac{5\sqrt{2}}{4}\)
\(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-\left(xy\right)^2\left(x+y\right)=\frac{3}{2}.\frac{5\sqrt{2}}{4}-\frac{1}{16}.\sqrt{2}=\frac{29\sqrt{2}}{16}\)
Vậy \(S=\frac{29\sqrt{2}}{16}\)
1) \(x^2+y=y^2+x\Leftrightarrow x^2-y^2-\left(x-y\right)=0\Leftrightarrow\left(x-y\right)\left(x+y-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=x\\y=1-x\end{cases}}\). Vì x,y là hai số khác nhau nên ta loại trường hợp x = y. Vậy ta có y = x-1.
\(P=\frac{x^2+\left(1-x\right)^2+x\left(1-x\right)}{x\left(1-x\right)-1}=\frac{x^2+x^2-2x+1-x^2+x}{-x^2+x-1}\)
\(=\frac{x^2-x+1}{-\left(x^2-x+1\right)}=-1\)
\(S=x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^2\left(x+y\right)\)\(=\left(x^2+y^2\right)\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]-x^2y^2\left(x+y\right)\)
\(=\left[\left(\frac{\sqrt{2}+1}{2}\right)^2+\left(\frac{\sqrt{2}-1}{2}\right)^2\right]\)\(\cdot\left[\left(\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2}-1}{2}\right)^3-3\frac{\sqrt{2}+1}{2}\frac{\sqrt{2}-1}{2}\left(\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2-1}}{2}\right)\right]\)\(-\left(\frac{\sqrt{2}+1}{2}\right)^2\left(\frac{\sqrt{2}-1}{2}\right)^2\left(\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2}-1}{2}\right)\)
\(=\frac{\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}-1\right)^2}{4}\cdot\left[\left(\sqrt{2}\right)^3-\frac{3}{4}\sqrt{2}\right]\)\(-\frac{\left(\sqrt{2}+1\right)^2\left(\sqrt{2}-1\right)^2}{4\cdot4}\sqrt{2}\)
\(=\frac{2+2\sqrt{2}+1+2-2\sqrt{2}+1}{4}\left(2\sqrt{2}-\frac{3\sqrt{2}}{4}\right)-\frac{1}{16}\sqrt{2}\)
\(=\frac{6}{4}\cdot\frac{5\sqrt{2}}{4}-\frac{\sqrt{2}}{16}=\frac{30\sqrt{2}-\sqrt{2}}{16}=\frac{29\sqrt{2}}{16}\)