1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)

\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)

\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)

\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)

\(=-8\sqrt{3}\)

2) \(A=\sqrt{12-4x}\) có nghĩa khi:

\(12-4x\ge0\)

\(\Leftrightarrow4x\le12\)

\(\Leftrightarrow x\le\dfrac{12}{4}\)

\(\Leftrightarrow x\le3\)

3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)

\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}\right)^2-1^2}\)

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

\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)

a,\(A=2\sqrt{x^2+x+\dfrac{1}{2}}=2\sqrt{x^2+x+\dfrac{1}{4}+\dfrac{1}{4}}=2\sqrt{\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{4}}\)

\(=\sqrt{4\left(x+\dfrac{1}{2}\right)^2+1}\ge1\) dấu"=" xảy ra<=>x=-1/2

\(B=\sqrt{2\left(x^2-2x+\dfrac{5}{2}\right)}=\sqrt{2\left[x^2-2x+1+\dfrac{3}{2}\right]}\)

\(=\sqrt{2\left(x-1\right)^2+3}\ge\sqrt{3}\) dấu"=" xảy ra<=>x=1

\(C=\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\ge\dfrac{-2}{-\sqrt{2}}=\sqrt{2}\) dấu"=" xảy ra<=>x=1

\(D=x-2\sqrt{x+2}\ge-2\) dấu"=" xảy ra<=>x=-2

d)D=\(x-2\sqrt{x+2}=\left(x+2\right)-2\sqrt{x+2}+1-3\)

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

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

Bài 1: 

Ta có: \(D=\sqrt{16x^4}-2x^2+1\)

\(=4x^2-2x^2+1\)

\(=2x^2+1\)

a: \(P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\)

\(M=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

b: \(A=\dfrac{-3x+4x+7}{\sqrt{x}+3}=\dfrac{x+7}{\sqrt{x}+3}=\dfrac{x-9+16}{\sqrt{x}+3}\)

=>\(A=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\sqrt{16}-6=2\)

Dấu = xảy ra khi 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 \(x=0\)

1.

Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:
$A=|x+2|+|x+3|=|x+2|+|-x-3|\geq |x+2-x-3|=1$

Vậy GTNN của $A$ là $1$. Giá trị này đạt tại $(x+2)(-x-3)\geq 0$

$\Leftrightarrow (x+2)(x+3)\leq 0$

$\Leftrightarrow -3\leq x\leq -2$

2. ĐKXĐ: $x\geq 1$

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

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

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

Vậy gtnn của $B$ là $2$. Giá trị này đạt tại $(\sqrt{x-1}+1)(1-\sqrt{x-1})\geq 0$

$\Leftrightarrow 1-\sqrt{x-1}\geq 0$

$\Leftrightarrow 0\leq x\leq 2$

3.

$C\sqrt{2}=\sqrt{4x+2\sqrt{4x-1}}+\sqrt{4x+2\sqrt{4x-1}}$

$=2\sqrt{(4x-1)+2\sqrt{4x-1}+1}=2\sqrt{(\sqrt{4x-1}+1)^2}$
$=2|\sqrt{4x-1}+1|$

Vì $\sqrt{4x-1}\geq 0$ nên $|\sqrt{4x-1}+1|\geq 1$

$\Rightarrow C\sqrt{2}\geq 2$

$\Rightarrow C\geq \sqrt{2}$

Vậy $C_{\min}=\sqrt{2}$. Giá trị này đạt tại $x=\frac{1}{4}$

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

\(\Leftrightarrow x^2+2x+1=x\sqrt[]{x^2+1}+2\sqrt[]{x^2+1}\left(x\ge-2\right)\)

\(\Leftrightarrow\left(x^2+2x+1\right)^2=\left(x\sqrt[]{x^2+1}+2\sqrt[]{x^2+1}\right)^2\)

\(\Leftrightarrow x^4+4x^2+1+4x^3+2x^2+4x=x^2\left(x^2+1\right)+4\left(x^2+1\right)+4x\left(x^2+1\right)\)

\(\Leftrightarrow x^4+4x^3+6x^2+4x+1=x^4+x^2+4x^2+4+4x^3+4\)

\(\Leftrightarrow x^4+4x^3+6x^2+4x+1=x^4+4x^3+5x^2+4x+4\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow x=\pm\sqrt[]{3}\left(Tm.x\ge-2\right)\)

Vậy nghiệm của phương trình \(\left(1\right)\) là \(x=\pm\sqrt[]{3}\)

2) \(P=\sqrt[]{x^2-2x+13}+4\sqrt[]{x-3}\)

Ta có : 

\(\sqrt[]{x^2-2x+13}=\sqrt[]{x^2-2x+1+12}=\sqrt[]{\left(x-1\right)^2+12}\ge\sqrt[]{12}=2\sqrt[]{3},\forall x\in R\)

\(4\sqrt[]{x-3}\ge0,\forall x\ge3\)

\(\Rightarrow P=\sqrt[]{x^2-2x+13}+4\sqrt[]{x-3}\ge\sqrt[]{4+12}+0=4\left(khi.x=3\right),\forall x\ge3\)

Vậy \(Min\left(P\right)=4\left(tại.x=3\right)\)

ta có :

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

Dấu bằng xảy ra khi : \(\left(x+1\right)\left(x+2\right)\le0\Leftrightarrow-2\le x\le-1\)