Bài 1:

1: \(P=\left(\frac{2\sqrt{x}+1}{x+\sqrt{x}}-\frac{1-\sqrt{x}}{\sqrt{x}}\right):\left(1+\frac{2}{\sqrt{x}}\right)\)

\(=\left(\frac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\sqrt{x}}\right):\frac{\sqrt{x}+2}{\sqrt{x}}\)

\(=\frac{2\sqrt{x}+1+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}}{\sqrt{x}+2}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)

2: \(x=2019-2\cdot\sqrt{2018}=\left(\sqrt{2018}-1\right)^2\)

\(P=\frac{\sqrt{x}}{\sqrt{x}+1}\)

\(=\frac{\sqrt{\left(\sqrt{2018}-1\right)^2}}{\sqrt{\left(\sqrt{2018}-1\right)^2}+1}=\frac{\sqrt{2018}-1}{\sqrt{2018}-1+1}=\frac{\sqrt{2018}-1}{\sqrt{2018}}\)

\(=\frac{2018-\sqrt{2018}}{2018}\)

\(A=\left(\frac{x-1}{\sqrt{x}-1}+\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}\right).\frac{1}{2\sqrt{x}}=\left[\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x-1}}+\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\right].\frac{1}{2\sqrt{x}}\)

\(A=2\left(\sqrt{x}+1\right).\frac{1}{2\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}}>1=\sqrt{\frac{2019}{2019}}>\sqrt{\frac{2018}{2019}}\) ( đpcm ) 

... 

1/x + 1/y = 1/2018

<=> 1/x = 1/2018 - 1/y = (y - 2018)/(2018y) 

<=> x = 2018y/(y - 2018) 

=> x + y = 2018y/(y - 2018) + y = y^2/(y - 2018) 

=> x - 2018 = 2018y/(y - 2018) - 2018 = 2018^2/(y - 2018) 

=> P = 1

a, \(\frac{\sqrt{2}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\)-\(\frac{3\left(1+\sqrt{3}\right)}{1+\sqrt{3}}\)

=\(\sqrt{2}-3\)

b,X=\(\sqrt{2019}+\sqrt{2018}\)

(Khử mẫu,nhân tử&mẫu vs\(\sqrt{2019}+\sqrt{2018}\))

Y=\(\sqrt{2018}+\sqrt{2017}\)

(Khử mẫu,nhân tử&mẫu vs\(\sqrt{2018}+\sqrt{2017}\))

So sánh:X & Y<=>X-\(\sqrt{2018}\)&Y-\(\sqrt{2018}\)(Trừ hai vế cho \(\sqrt{2018}\)) <=>\(\sqrt{2019}\)&\(\sqrt{2017}\)

Có:2019>2017

=>\(\sqrt{2019}>\sqrt{2017}\)

=>X>Y

Câu b, mk ko bt có lm đúng ko?