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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\left\{{}\begin{matrix}\left(x+\sqrt{2017+x^2}\right)\left(\sqrt{2017+x^2}-x\right)=2017\\\left(x+\sqrt{2017+x^2}\right)\left(y+\sqrt{2017+y^2}\right)=2017\end{matrix}\right.\)
\(\Rightarrow\sqrt{2017+x^2}-x=y+\sqrt{2017+y^2}\)
\(\Leftrightarrow x+y=\sqrt{2017+x^2}-\sqrt{2017+y^2}\left(1\right)\)
\(\left\{{}\begin{matrix}\left(y+\sqrt{2017+y^2}\right)\left(\sqrt{2017+y^2}-y\right)=2017\\\left(y+\sqrt{2017+y^2}\right)\left(x+\sqrt{2017+x^2}\right)=2017\end{matrix}\right.\)
\(\Rightarrow\sqrt{2017+y^2}-y=x+\sqrt{2017+x^2}\)
\(\Leftrightarrow x+y=\sqrt{2017+y^2}-\sqrt{2017+x^2}\left(2\right)\)
Lấy (1) + (2) \(\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\Leftrightarrow x=-y\)
\(T=x^{2017}+y^{2017}=-y^{2017}+y^{2017}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(P^2\)=\(\dfrac{x+y}{x+y-4034+2\sqrt{\left(x-2017\right)\left(y-2017\right)}}\)
\(P^2\)=\(\dfrac{x+y}{x+y-4034+2\sqrt{xy-2017\left(x+y\right)+2017^2}}\)
Mà \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2017}\)
Suy ra xy=2017(x+y)
Suy ra \(P^2=\dfrac{x+y}{x+y-4034+2\sqrt{2017\left(x+y\right)-2017\left(x+y\right)+2017^2}}\)
\(P^2=\dfrac{x+y}{x+y-4034+2\sqrt{2017^2}}\)
\(P^2=\dfrac{x+y}{x+y-4034+4034}=\dfrac{x+y}{x+y}=1\)
Vậy P=1
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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(x+\sqrt[]{x^2+2017}\right)\left(x-\sqrt[]{x^2+2017}\right)=x^2-x^2-2017=-2017\)
Mà \(\left(x+\sqrt[]{x^2+2017}\right)\left(y+\sqrt[]{y^2+2017}=2017\right)\)
Nên \(\sqrt[]{x^2+2017}-x=y+\sqrt[]{y^2+2017}\) (1)
Chứng minh tương tư: \(\sqrt[]{y^2+2017}-y=x+\sqrt[]{x^2+2017}\) (2)
Cộng hai vế của (1) và (2) \(\Rightarrow-x-y=x+y\Rightarrow-2\left(x+y\right)=0\Rightarrow S=x+y=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
từ pt đã cho
=> (x - \(\sqrt{x^2+1}\)) (x+\(\sqrt{x^2+1}\)) (y+\(\sqrt{y^2+1}\))
= x - \(\sqrt{x^2-1}\) (x-\(\sqrt{x^2+1}\) luôn khác 0 tự cm)
thu gọn 2 vế
=> - y - \(\sqrt{y^2+1}\) = x -\(\sqrt{x^2+1}\) (1)
tương tự khi nhân 2 vế pt đầu với y - \(\sqrt{y^2+1}\)
=> - x - \(\sqrt{x^2+1}\) = y - \(\sqrt{y^2+1}\) (2)
cộng vế với vế (1) và (2)
=> - 2 (x+y) = 0 => x+y = 0 => x = - y
=>A = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(2017=a\)
\(A=\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2a+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2\left(a+1\right)\cdot\dfrac{a}{a+1}+\left(\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1-\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\left|a+1-\dfrac{a}{a+1}\right|+\dfrac{a}{a+1}\)
Ta có \(\dfrac{a}{a+1}< 1\Leftrightarrow a+1-\dfrac{a}{a+1}>0\)
\(\Leftrightarrow A=a+1-\dfrac{a}{a+1}+\dfrac{a}{a+1}=a+1=2018\)
https://hoc24.vn/hoi-dap/question/258448.html
x + y = 0
=> x = - y (1)
Thay (1) vào P, ta có:
P = x2017 - x2017 + 2017 = 2017