\(\left(\dfrac{x+1}{\sqrt{x}+1}+\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right):\dfrac{\sqrt{x}}{x+2\sqrt{x}+1}=\left(\dfrac{x\sqrt{x}+\sqrt{x}}{x+\sqrt{x}}+\dfrac{1}{x+\sqrt{x}}-\dfrac{\sqrt{x}+1}{x}\right):\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)^2}=\dfrac{x\sqrt{x}+\sqrt{x}}{x+\sqrt{x}}:\dfrac{\sqrt{x}}{\cdot\left(\sqrt{x}+1\right)^2}=\dfrac{\sqrt{x}\left(x+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x+1}\right)}=\dfrac{x+1}{\sqrt{x}+1}:\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(x+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x+1}\right)}{\left(\sqrt{x}+1\right)\sqrt{x}}=\dfrac{\left(x+1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{x\sqrt{x}+x+\sqrt{x}+1}{\sqrt{x}}\)
\(=x+\sqrt{x}+1+\dfrac{1}{\sqrt{x}}\ge2017+\sqrt{2017}\Leftrightarrow x+\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2016+\sqrt{2017}\Leftrightarrow x+\sqrt{x}+\dfrac{1}{\sqrt{x}}-2016-\sqrt{2017}\ge0\)
Bài toán sắp hoàn thành rồi đấy cậu giải tiếp nhé! =))