\(C=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\)
\(=\dfrac{x\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x-1}{\sqrt{x}+1}\)
\(=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
Đặt A=C-1
=>\(A=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\dfrac{1}{\sqrt{x}-1}\)
Thay \(x=2020+2\sqrt{2019}\) vào A, ta được:
\(A=\dfrac{1}{\sqrt{\left(2020+2\sqrt{2019}\right)}-1}\)
\(=\dfrac{1}{\sqrt{\left(\sqrt{2019}+1\right)^2}-1}=\dfrac{1}{\sqrt{2019}+1-1}=\dfrac{1}{\sqrt{2019}}=\dfrac{\sqrt{2019}}{2019}\)
