a) 

Do: \(y=\sqrt{x+2}\)

<=> \(y^2=x+2\)

<=> \(x=y^2-2\)

Khi đó: \(A=y^2-2-2y\)

Vậy \(A=y^2-2y-2\)

b) 

\(A=y^2-2y-2\left(cmt\right)\)

\(A=\left(y^2-2y+1\right)-3\)

\(A=\left(y-1\right)^2-3\)

Do \(\left(y-1\right)^2\ge0\forall y\)

=> \(\left(y-1\right)^2-3\ge-3\)

=> \(A\ge-3\)

Vậy A MIN = -3 <=> \(\left(y-1\right)^2=0\)

<=> \(y=1\)

Do: \(y=\sqrt{x+2}\)

<=> \(\sqrt{x+2}=1\)

<=> \(x+2=1\)

<=> \(x=-1\)

Bài 5: 

a: Thay \(x=4+2\sqrt{3}\) vào E, ta được:

\(E=\dfrac{\sqrt{3}+1-1}{\sqrt{3}+1-3}=\dfrac{\sqrt{3}}{\sqrt{3}-2}=-3-2\sqrt{3}\)

b: Để E<1 thì E-1<0

\(\Leftrightarrow\dfrac{\sqrt{x}-1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\sqrt{x}-3< 0\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)

c: Để E nguyên thì \(4⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{-2;1;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;5;7\right\}\)

hay \(x\in\left\{16;25;49\right\}\)

Câu 2:
a) Ta có \(x=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)

Thay \(x=\sqrt{3}-1\) vào \(B\), ta được

\(B=\dfrac{\sqrt{3}-1-2}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-3}{\sqrt{3}}=1-\sqrt{3}\)

b) Để \(B\) âm thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\) mà \(\sqrt{x}+1\ge1>0\forall x\) \(\Rightarrow\sqrt{x}-2< 0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)

c) Ta có \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}\)

Với mọi \(x\ge0\) thì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{3}{\sqrt{x}+1}\le3\Rightarrow B=1-\dfrac{3}{\sqrt{x}+1}\ge-2\)

Dấu "=" xảy ra khi \(\sqrt{x}+1=1\Leftrightarrow x=0\)

Vậy \(B_{min}=-2\) khi \(x=0\)

\(A=\sqrt{x}+\dfrac{2}{\sqrt{x}}\ge2\cdot\sqrt{\sqrt{x}\cdot\dfrac{2}{\sqrt{x}}}=2\sqrt{2}\)

Dấu '=' xảy ra khi \(\sqrt{x}\cdot\sqrt{x}=2\)

hay \(x=2\)

Áp dụng BĐT BSC và BĐT \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\):

\(A=x\sqrt{y+1}+y\sqrt{x+1}\)

\(\Rightarrow A^2=\left(x\sqrt{y+1}+y\sqrt{x+1}\right)^2\)

\(\le\left(x^2+y^2\right)\left(x+y+2\right)\)

\(\le\left(x^2+y^2\right)\left[\sqrt{2\left(x^2+y^2\right)}+2\right]=\sqrt{2}+2\)

\(\Rightarrow-\sqrt{\sqrt{2}+2}\le A\le\sqrt{\sqrt{2}+2}\)

\(\Rightarrow minA=\sqrt{\sqrt{2}+2}\Leftrightarrow x=y=-\dfrac{1}{\sqrt{2}}\)

Lời giải:
a.

\(A=\frac{\sqrt{x}(\sqrt{x^3}-1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}+\frac{2(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1}\)

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

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

b.

$A=x-\sqrt{x}+1=(x-\sqrt{x}+\frac{1}{4})+\frac{3}{4}$

$=(\sqrt{x}-\frac{1}{2})^2+\frac{3}{4}\geq 0+\frac{3}{4}=\frac{3}{4}$

$\Rightarrow A_{\min}=\frac{3}{4}$

Giá trị này đạt tại $\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}$