If x>0, y≥1 and a= 2xy/y^2+1 , the the value of 
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![x>0,\ \ y \ge 1,\ \ \ a=\frac{2xy}{y^2+1}\\\\x+a=x[1+\frac{2y}{y^2+1}]=x[\frac{(1+y)^2}{y^2+1}]\\\\x-a=x[1-\frac{2y}{y^2+1}]=x[\frac{(1-y)^2}{y^2+1}]\\\\x^2-a^2=\frac{x^2(1+y)^2(1-y)^2}{(y^2+1)^2}\\\\\frac{(\sqrt{x+a}-\sqrt{x-a})}{(\sqrt{x+a}+\sqrt{x-a})}\\\\=\frac{(\sqrt{x+a}-\sqrt{x-a})(\sqrt{x+a}-\sqrt{x-a})}{(\sqrt{x+a}+\sqrt{x-a})(\sqrt{x+a}-\sqrt{x-a})}\\\\=\frac{x+a+x-a-2\sqrt{(x+a)(x-a)}}{x+a-(x-a)}\\\\=\frac{1}{2a}[2x-2\sqrt{x^2-a^2}]\\\\=\frac{x}{a}[1-\frac{(1-y^2)}{1+y^2}]](https://tex.z-dn.net/?f=x%26gt%3B0%2C%5C+%5C+y+%5Cge+1%2C%5C+%5C+%5C+a%3D%5Cfrac%7B2xy%7D%7By%5E2%2B1%7D%5C%5C%5C%5Cx%2Ba%3Dx%5B1%2B%5Cfrac%7B2y%7D%7By%5E2%2B1%7D%5D%3Dx%5B%5Cfrac%7B%281%2By%29%5E2%7D%7By%5E2%2B1%7D%5D%5C%5C%5C%5Cx-a%3Dx%5B1-%5Cfrac%7B2y%7D%7By%5E2%2B1%7D%5D%3Dx%5B%5Cfrac%7B%281-y%29%5E2%7D%7By%5E2%2B1%7D%5D%5C%5C%5C%5Cx%5E2-a%5E2%3D%5Cfrac%7Bx%5E2%281%2By%29%5E2%281-y%29%5E2%7D%7B%28y%5E2%2B1%29%5E2%7D%5C%5C%5C%5C%5Cfrac%7B%28%5Csqrt%7Bx%2Ba%7D-%5Csqrt%7Bx-a%7D%29%7D%7B%28%5Csqrt%7Bx%2Ba%7D%2B%5Csqrt%7Bx-a%7D%29%7D%5C%5C%5C%5C%3D%5Cfrac%7B%28%5Csqrt%7Bx%2Ba%7D-%5Csqrt%7Bx-a%7D%29%28%5Csqrt%7Bx%2Ba%7D-%5Csqrt%7Bx-a%7D%29%7D%7B%28%5Csqrt%7Bx%2Ba%7D%2B%5Csqrt%7Bx-a%7D%29%28%5Csqrt%7Bx%2Ba%7D-%5Csqrt%7Bx-a%7D%29%7D%5C%5C%5C%5C%3D%5Cfrac%7Bx%2Ba%2Bx-a-2%5Csqrt%7B%28x%2Ba%29%28x-a%29%7D%7D%7Bx%2Ba-%28x-a%29%7D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2a%7D%5B2x-2%5Csqrt%7Bx%5E2-a%5E2%7D%5D%5C%5C%5C%5C%3D%5Cfrac%7Bx%7D%7Ba%7D%5B1-%5Cfrac%7B%281-y%5E2%29%7D%7B1%2By%5E2%7D%5D "x>0,\ \ y \ge 1,\ \ \ a=\frac{2xy}{y^2+1}\\\\x+a=x[1+\frac{2y}{y^2+1}]=x[\frac{(1+y)^2}{y^2+1}]\\\\x-a=x[1-\frac{2y}{y^2+1}]=x[\frac{(1-y)^2}{y^2+1}]\\\\x^2-a^2=\frac{x^2(1+y)^2(1-y)^2}{(y^2+1)^2}\\\\\frac{(\sqrt{x+a}-\sqrt{x-a})}{(\sqrt{x+a}+\sqrt{x-a})}\\\\=\frac{(\sqrt{x+a}-\sqrt{x-a})(\sqrt{x+a}-\sqrt{x-a})}{(\sqrt{x+a}+\sqrt{x-a})(\sqrt{x+a}-\sqrt{x-a})}\\\\=\frac{x+a+x-a-2\sqrt{(x+a)(x-a)}}{x+a-(x-a)}\\\\=\frac{1}{2a}[2x-2\sqrt{x^2-a^2}]\\\\=\frac{x}{a}[1-\frac{(1-y^2)}{1+y^2}]")
![=\frac{x}{a}[\frac{1+y^2-1+y^2}{1+y^2}]\\\\=\frac{1}{a}\frac{2xy^2}{(1+y^2)}\\\\=\frac{1}{a}a*y\\\\=y](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bx%7D%7Ba%7D%5B%5Cfrac%7B1%2By%5E2-1%2By%5E2%7D%7B1%2By%5E2%7D%5D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7Ba%7D%5Cfrac%7B2xy%5E2%7D%7B%281%2By%5E2%29%7D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7Ba%7Da%2Ay%5C%5C%5C%5C%3Dy "=\frac{x}{a}[\frac{1+y^2-1+y^2}{1+y^2}]\\\\=\frac{1}{a}\frac{2xy^2}{(1+y^2)}\\\\=\frac{1}{a}a*y\\\\=y")
So answer is y >= 1.
So answer is y >= 1.
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