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

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

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Lời giải:

ĐK: $x\geq 0; x\neq 1$

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

\(=\frac{x+1-2\sqrt{x}}{(\sqrt{x}-1)(x+1)}:\frac{\sqrt{x}+1}{x+1}=\frac{(\sqrt{x}-1)^2}{(x+1)(\sqrt{x}-1)}.\frac{x+1}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

b)

Với $B=\frac{\sqrt{x}-1}{\sqrt{x}+1}$ thôi thì $0< B< 2$ là không đúng. Bạn cho thử $x=0,5$ sẽ thấy.

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

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

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

2) Xét hiệu  \(A-\frac{1}{3}=\frac{2\sqrt{x}}{x^2+x+1}-\frac{1}{3}=\frac{6\sqrt{x}-\left(x^2+x+1\right)}{3\left(x^2+x+1\right)}\)

Mẫu luôn > 0

Tử chưa chắc < 0 .Ví dụ lấy x = 2 thì tử > 0 => Không khẳng định được A < 1/3

Chứng tỏ 0<Q<2 nha

\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+1=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}\)

\(P+1=\frac{x^2+x+1}{x+\sqrt{x}+1}=\frac{x^2+2x+1-x}{x+\sqrt{x}+1}=\frac{\left(x+\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x+\sqrt{x}+1}=x-\sqrt{x}+1\ge\frac{3}{4}\)

a) \(ĐKXĐ:x>1\)

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

\(=\frac{\left(\sqrt{x}\right)^4-\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}.\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}.\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)

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

\(=\sqrt{x}.\left(\sqrt{x}-1\right)+1=x-\sqrt{x}+1\)

b) Ta có: \(P=x-\sqrt{x}+1=x-2.\frac{1}{2}.\sqrt{x}+\frac{1}{4}+\frac{3}{4}\)

\(=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\left(\forall x>0\right)\)\(\Rightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu " = " xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\)\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)\(\Leftrightarrow x=\frac{1}{4}\)

Vậy \(minP=\frac{3}{4}\Leftrightarrow x=\frac{1}{4}\)