b, ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

Ta có : \(B=\dfrac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{x+2}{\sqrt{x}}\)

\(=\dfrac{x+2\sqrt{x}+4}{\sqrt{x}}-\dfrac{x-2\sqrt{x}+4}{\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)

\(=\dfrac{x+2\sqrt{x}+4-x+2\sqrt{x}-4+x+2}{\sqrt{x}}\)

\(=\dfrac{x+4\sqrt{x}+2}{\sqrt{x}}\)

b) Ta có: \(B=\dfrac{x\sqrt{x}-8}{x-2\sqrt{x}}-\dfrac{x\sqrt{x}+8}{x+2\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)

\(=\dfrac{x+2\sqrt{x}+4}{\sqrt{x}}-\dfrac{x-2\sqrt{x}+4}{\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)

\(=\dfrac{4\sqrt{x}+x+2}{\sqrt{x}}\)

c) Ta có: \(C=\dfrac{1}{\sqrt{x}+2}-\dfrac{5}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}\)

\(=\dfrac{\sqrt{x}-3-5+\left(x-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)

1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

Để \(A=-\dfrac{1}{\sqrt{x}}\) thì \(x+\sqrt{x}=-\sqrt{x}+3\)

\(\Leftrightarrow x+2\sqrt{x}-3=0\)

\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow x=1\left(nhận\right)\)

2: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

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

\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)

\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)

1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)

Ta có: \(A=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

\(1,A=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\left(x\ge0;x\ne4;x\ne9\right)\\ 2,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow0\le x< 9\)

1:

a: \(A=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

căn x+1>=1

=>2/căn x+1<=2

=>-2/căn x+1>=-2

=>A>=-2+1=-1

Dấu = xảy ra khi x=0

b: 

*Rút gọn

Ta có: \(C=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

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

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

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

Ta có: \(C=x-\sqrt{x}+1\)

\(=x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\sqrt{x}=\dfrac{1}{2}\)

hay \(x=\dfrac{1}{4}\)

\(C=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\left(x>0;x\ne1\right)\)

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

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

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

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

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

Vậy  \(C_{min}=\dfrac{3}{4}\)

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

Áp dụng AM-GM có: \(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\)

Dấu "=" xảy ra khi x=1 (ktm đk)

Suy ra dấu bằng ko xảy ra \(\Rightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}-1>2-1=1\)

\(\Rightarrow\dfrac{2}{\sqrt{x}+\dfrac{1}{\sqrt{x}}-1}< 2\) 

\(\Rightarrow N< 2\) mà \(N>0\),\(N\) nguyên

\(\Rightarrow N=1\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{5}}{2}\\\sqrt{x}=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7+3\sqrt{5}}{2}\\x=\dfrac{7-3\sqrt{5}}{2}\end{matrix}\right.\) (tm)

Vậy...

\(\Rightarrow C=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\) * \(\Rightarrow C=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\) Dấu = xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

* Ta có \(N=\dfrac{2\sqrt{x}}{C}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}>0\left(1\right)\) 

Xét \(N-2=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}-2=\dfrac{2\sqrt{x}-2x+2\sqrt{x}-2}{x-\sqrt{x}+1}=\dfrac{-2x+4\sqrt{x}-2}{x-\sqrt{x}+1}=\dfrac{-2\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}< 0\left(dox\ne1\right)\Rightarrow N< 2\left(2\right)\) Từ (1) và (2) \(\Rightarrow0< N< 2\). Mà N nguyên nên N=1  \(\Rightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1\Rightarrow2\sqrt{x}=x-\sqrt{x}+1\Leftrightarrow x-3\sqrt{x}+1=0\)

\(\Delta=9-4=5\Rightarrow\) pt có 2 nghiệm phân biệt: \(x_1=\dfrac{\sqrt{5}+3}{2}\left(TM\right);x_2=\dfrac{3-\sqrt{5}}{2}\left(TM\right)\)

a: Ta có: \(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

a) Điều kiện: \(x\ge0;x\ne1;x\ne\dfrac{1}{4}\)\(E=\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt[]{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)

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

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

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

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

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

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

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

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

\(E=\dfrac{x+\sqrt{x}}{x+\sqrt{x}+1}\)

b)Vì \(x\ge0\) nên \(x+\sqrt{x}\ge0\) và \(x+\sqrt{x}+1>0\)

Do đó: \(E\ge0\). Dấu "=" xảy ra \(\Leftrightarrow x=0\)

c)\(E\ge\dfrac{6}{7}\Leftrightarrow\dfrac{x+\sqrt{x}}{x+\sqrt{x}+1}\ge\dfrac{6}{7}\Leftrightarrow7x+7\sqrt{x}\ge6x+6\sqrt{x}+6\)

                \(\Leftrightarrow x+\sqrt{x}-6\ge0\Leftrightarrow x-2\sqrt{x}+3\sqrt{x}-6\ge0\)

                 \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ge0\)

                  \(\Leftrightarrow\sqrt{x}-2\ge0\Leftrightarrow\sqrt{x}\ge2\Leftrightarrow x\ge4\)

Bài 1:

$\sqrt{x-4}-2$
ĐKXĐ: $x\geq 4$
Ta thấy $\sqrt{x-4}\geq 0$ với mọi $x\geq 4$
$\Rightarrow \sqrt{x-4}-2\geq 0-2=-2$
Vậy gtnn của biểu thức là $-2$. Giá trị này đạt được tại $x-4=0$

$\Leftrightarrow x=4$

Bài 2: $x-\sqrt{x}$

ĐKXĐ: $x\geq 0$

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

$\geq 0-\frac{1}{4}=\frac{-1}{4}$
Vậy gtnn của biểu thức là $\frac{-1}{4}$. Giá trị này đạt được khi $\sqrt{x}-\frac{1}{2}=0$

$\Leftrightarrow x=\frac{1}{4}$

Bài 3:

$x-4\sqrt{x}+10$

ĐKXĐ: $x\geq 0$

Ta có: $x-4\sqrt{x}+10=(x-4\sqrt{x}+4)+6=(\sqrt{x}-2)^2+6\geq 0+6=6$

Vậy gtnn của biểu thức là $6$. Giá trị này đạt được khi $\sqrt{x}-2=0\Leftrightarrow x=4$