Bài 3:

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

\(=3+2\sqrt{2}-\sqrt{2}=3+\sqrt{2}\)

b: \(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\)

\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\left(\sqrt{ab}-b\right)}{\left(a+\sqrt{b}\right)^2}\)

\(=\dfrac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=\dfrac{b}{a+\sqrt{b}}\)

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

a) \(x\sqrt{x}+\sqrt{x}-x-1\) 

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

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

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

b) \(\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)

\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)

\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

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

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

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

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

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

26: \(x^2-\sqrt{x}+x-1\)

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

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

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

25: Ta có: \(-6x+7\sqrt{x}-2\)

\(=-6x+3\sqrt{x}+4\sqrt{x}-2\)

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

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

27: Ta có: \(2a-5\sqrt{ab}+3b\)

\(=2a-2\sqrt{ab}-3\sqrt{ab}+3b\)

\(=2\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\left(\sqrt{a}-\sqrt{b}\right)\left(2\sqrt{a}-3\sqrt{b}\right)\)

28: Ta có: \(\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)

\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)

\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)

a: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}+4}{x\sqrt{x}-3x+2\sqrt{x}}-\dfrac{3\sqrt{x}+3}{-x+\sqrt{x}+2}\right):\left(\dfrac{x-\sqrt{x}-6}{x-3\sqrt{x}}-\dfrac{x-2\sqrt{x}}{x-4\sqrt{x}+4}\right)+\sqrt{x}\)

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

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

\(=\dfrac{4\left(\sqrt{x}-1\right)}{-4}+\sqrt{x}\)

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

=-1

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

\(=\dfrac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x\sqrt{x}-x-4\sqrt{x}-6-2x+12\sqrt{x}-18}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x\left(\sqrt{x}-3\right)+8\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x+8}{\sqrt{x}+1}\)

Lời giải:
ĐKXĐ: $x\geq 0; x\neq 9$

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

\(=\frac{x\sqrt{x}-3x+8\sqrt{x}-24}{(\sqrt{x}+1)(\sqrt{x}-3)}=\frac{(\sqrt{x}-3)(x+8)}{(\sqrt{x}+1)(\sqrt{x}-3)}=\frac{x+8}{\sqrt{x}+1}\)

b.

\(14-6\sqrt{5}=(3-\sqrt{5})^2\Rightarrow \sqrt{x}=3-\sqrt{5}\)

\(A=\frac{14-6\sqrt{5}+8}{3-\sqrt{5}+1}=\frac{22-6\sqrt{5}}{4-\sqrt{5}}=\frac{58-2\sqrt{5}}{11}\)

c. 

Áp dụng BĐT Cô-si:
$x+4\geq 4\sqrt{x}\Rightarrow x+8\geq 4(\sqrt{x}+1)$

$\Rightarrow A=\frac{x+8}{\sqrt{x}+1}\geq 4$

Vậy $A_{\min}=4$. Giá trị này đạt tại $x=4$

\(\text{a)}x\sqrt{x}+\sqrt{x}-x-1\)

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

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

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

\(\text{b)}\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)

\(=\left(\sqrt{ab}+2\sqrt{a}\right)+\left(3\sqrt{b}+6\right)\)

\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)

\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)

\(\text{c)}\left(1+\sqrt{x}\right)^2-4\sqrt{x}\)

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

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

\(\text{d)}\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)

\(=\left(\sqrt{ab}-\sqrt{a}\right)-\left(\sqrt{b}-1\right)\)

\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)

\(=\left(\sqrt{b}-1\right)\left(\sqrt{a}-1\right)\)

\(\text{e)}a+\sqrt{a}+2\sqrt{ab}+2\sqrt{b}\)

\(=\left(a+\sqrt{a}\right)+\left(2\sqrt{ab}+2\sqrt{b}\right)\)

\(=\left[\left(\sqrt{a}\right)^2+\sqrt{a}\right]+\left(2\sqrt{ab}+2\sqrt{b}\right)\)

\(=\sqrt{a}\left(\sqrt{a}+1\right)+2\sqrt{b}\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\sqrt{b}\right)\)

\(\text{f)}x-2\sqrt{x-1}-a^2\)

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

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

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

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