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 \(x=0\)

ĐKXĐ : \(x\ne0;x\ne\pm1\)

a) Bạn ghi lại rõ đề.

b) \(B=\dfrac{x-1}{x+1}+\dfrac{3x-x^2}{x^2-1}=\dfrac{x-1}{x+1}+\dfrac{3x-x^2}{\left(x-1\right).\left(x+1\right)}\)

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

c) \(P=A.B=\dfrac{x^2+x-2}{x.\left(x-1\right)}=\dfrac{\left(x-1\right).\left(x+2\right)}{x\left(x-1\right)}=\dfrac{x+2}{x}=1+\dfrac{2}{x}\)

Không tồn tại Min P \(\forall x\inℝ\)

a.

\(A=x^2+\dfrac{2021}{x}=x^2+\dfrac{2021}{2x}+\dfrac{2021}{2x}\ge3\sqrt[3]{\dfrac{2021^2}{4x^2}}=3\sqrt[3]{\dfrac{2021^2}{4}}\)

Dấu "=" xảy ra khi \(x=\sqrt[3]{\dfrac{2021}{3}}\)

b.

\(B=4\left(x-1\right)+\dfrac{25}{x-1}+4\ge2\sqrt{\dfrac{100\left(x-1\right)}{x-1}}+4=24\)

Dấu "=" xảy ra khi \(x=\dfrac{7}{2}\)

c.

\(C=3x+\dfrac{16}{x^3}=x+x+x+\dfrac{16}{x^3}\ge4\sqrt[4]{\dfrac{16x^3}{x^3}}=8\)

\(A_{min}=8\) khi \(x=2\)

d.

\(D=x+\dfrac{1}{x}=\left(\dfrac{x}{4}+\dfrac{1}{x}\right)+\dfrac{3}{4}.x\ge2\sqrt{\dfrac{x}{4x}}+\dfrac{3}{4}.2=\dfrac{5}{2}\)

Dấu "=" xảy ra khi \(x=2\)

e.

\(E=\dfrac{9\left(x-2\right)+18}{2-x}+\dfrac{2}{x}=2\left(\dfrac{1}{x}+\dfrac{9}{2-x}\right)-9\ge\dfrac{2.\left(1+3\right)^2}{x+2-x}-9=7\)

\(E_{min}=7\) khi \(x=\dfrac{1}{5}\)

f.

\(F=\dfrac{3}{1-x}+\dfrac{4}{x}\ge\dfrac{\left(\sqrt{3}+2\right)^2}{1-x+x}=7+4\sqrt{3}\)

Dấu "=" xảy ra khi \(x=4-2\sqrt{3}\)

a: Để \(\dfrac{3x-2}{4}\) không nhỏ hơn \(\dfrac{3x+3}{6}\) thì \(\dfrac{3x-2}{4}>=\dfrac{3x+3}{6}\)

=>\(\dfrac{6\left(3x-2\right)}{24}>=\dfrac{4\left(3x+3\right)}{24}\)

=>18x-12>=12x+12

=>6x>=24

=>x>=4

b: Để \(\left(x+1\right)^2\) nhỏ hơn \(\left(x-1\right)^2\) thì \(\left(x+1\right)^2< \left(x-1\right)^2\)

=>\(x^2+2x+1< x^2-2x+1\)

=>4x<0

=>x<0

c: Để \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\) không lớn hơn \(\dfrac{x^2}{7}-\dfrac{2x-3}{5}\) thì

\(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}< =\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)

=>\(\dfrac{2x-3+5x\left(x-2\right)}{35}< =\dfrac{5x^2-7\cdot\left(2x-3\right)}{35}\)

=>\(2x-3+5x^2-10x< =5x^2-14x+21\)

=>-8x-3<=-14x+21

=>6x<=24

=>x<=4

b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{x^2-9}=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

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 $\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}$