Lời giải:

a.

 \(A=\frac{2(\sqrt{x}-4)-3(\sqrt{x}+4)}{(\sqrt{x}-4)(\sqrt{x}+4)}+\frac{2\sqrt{x}+16}{(\sqrt{x}-4)(\sqrt{x}+4)}=\frac{-\sqrt{x}-20}{(\sqrt{x}-4)(\sqrt{x}+4)}+\frac{2\sqrt{x}+16}{(\sqrt{x}-4)(\sqrt{x}+4)}\\ =\frac{\sqrt{x}-4}{(\sqrt{x}-4)(\sqrt{x}+4)}=\frac{1}{\sqrt{x}+4}\)

b. Khi $x=4-2\sqrt{3}=(\sqrt{3}-1)^2\Rightarrow \sqrt{x}=\sqrt{3}-1$

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

Lời giải:
a. ĐKXĐ: $x>0; x\neq 4$

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

b.

\(x=3+2\sqrt{2}=(\sqrt{2}+1)^2\Rightarrow \sqrt{x}=\sqrt{2}+1\)

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

c.

$M>0\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}}>0$

$\Leftrightarrow \sqrt{x}-2>0$

$\Leftrightarrow \sqrt{x}>2\Leftrightarrow x>4$

Kết hợp đkxđ suy ra $x>4$

\(a,\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}+2}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\left(dkxd:x\ge0;x\ne1\right)\)

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

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

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

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

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

\(b,\) Thay \(x=4-2\sqrt{3}\) vào biểu thức trên, ta được:

\(\sqrt{4-2\sqrt{3}}+1\)

\(=\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}+1\)

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

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

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

\(=\sqrt{3}\)

Vậy: ...

\(\text{#}Toru\)

\(a\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}+2}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\\ =\left(\dfrac{\sqrt{x}-1}{x-1}+\dfrac{x+\sqrt{x}+2}{x-1}\right).\sqrt{x}-1\\ =\dfrac{x+\sqrt{2}+1}{x-1}.\sqrt{x}-1\\ =\sqrt{x}+1\\ b,tacóx=4-2\sqrt{3}=\left(\sqrt{3}-\sqrt{1}\right)^2thãy=\sqrt{3}-\sqrt{1}vàobiểuthức,tađược\\ \sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}-1=\sqrt{3}-1-1=\sqrt{3}-2\)

\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)

__

Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)

\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)

a) ĐKXĐ: \(x\ge0;x\ne9;x\ne4\)

\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

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

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

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

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

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

b) Ta có M ϵ Z thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3}{\sqrt{x}-3}+\dfrac{4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)

Phải thuộc Z vậy:

4 ⋮ \(\sqrt{x}-3\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Mà: \(x\ge0,x\ne4,x\ne9\) nên \(\sqrt{x}-3\in\left\{1;2;-2;4\right\}\)

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

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

A=\(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)}+\frac{x}{\sqrt{x}+2}\right)\)

A=\(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\sqrt{x}+x}{\left(\sqrt{x}+2\right)}\right)\)

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

b, A >= \(\frac{1}{3\sqrt{x}}\)

=> \(\frac{1}{x+2\sqrt{x}}\) >= \(\frac{1}{3\sqrt{x}}\)

=> x <= -1 , x >= 4 (x khác 0)

√x√x−2−6√x−4x−4(x\(\ge\)0,x\(\ne\)4)

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

=\(\dfrac{x+2\sqrt{x}-6\sqrt{x}+4}{x-4}\)=\(\dfrac{x-4\sqrt{x}+4}{x-4}\)=\(\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}\)

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

b, với x=6-4\(\sqrt{2}\)=(2-\(\sqrt{2}\))^2 thay vào (1) ta được

\(\dfrac{\sqrt{\left(2-\sqrt{2}\right)}^2-2}{\sqrt{\left(2-\sqrt{2}\right)}^2+2}\)=\(\dfrac{2-\sqrt{2}-2}{2-\sqrt{2}+2}\)=\(\dfrac{-\sqrt{2}}{4-\sqrt{2}}\)=\(\dfrac{\sqrt{2}}{\sqrt{2}-4}\)

a)ĐKXĐ: x≠4;x≥0

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

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

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

b) thế x=\(6-4\sqrt{2}\) (thỏa mãn) vào bt ta đc:

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

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

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

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

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