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\(=\frac{x-2\sqrt{x}+y+2\sqrt{y}-2\sqrt{xy}+1}{x-2\sqrt{xy}+y-1}\)\(=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2-2\left(\sqrt{x}-\sqrt{y}\right)+1}{\left(\sqrt{x}-\sqrt{y}\right)^2-1}\)
\(=\frac{\left(\sqrt{x}-\sqrt{y}-1\right)^2}{\left(\sqrt{x}-\sqrt{y}+1\right)\left(\sqrt{x}-\sqrt{y}-1\right)}=\frac{\sqrt{x}-\sqrt{y}-1}{\sqrt{x}-\sqrt{y}+1}\)
Thay \(xy+yz+xz=1\) ta có: \(\hept{\begin{cases}1+x^2=xy+yz+xz+x^2=\left(x+z\right)\left(x+y\right)\\1+y^2=xy+yz+xz+y^2=\left(x+y\right)\left(y+z\right)\\1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\end{cases}}\)
\(\Rightarrow S=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+xz\right)=2\)
\(\sqrt{2000}\)=\(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(\Rightarrow2000=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+y^2\right)\left(1+x^2\right)}\)
=\(x^2y^2+1+x^2+y^2+x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(\Rightarrow x^2+y^2+2x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2000-1=1999\)
ma \(S^2=x^2\left(1+y^2\right)+y^2\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
=\(x^2+x^2y^2+y^2+x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
=\(x^2+y^2+2x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\) =\(1999\Rightarrow S=\sqrt{1999}\)
\(F=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y}{xy}\cdot\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\dfrac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y+2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}\cdot xy=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
bình phương là thấy ngay ,mà hình như nhầm chỗ 1-y2 =))
chỉ làm ik bạn <3 , mình làm k ra