Trắc nghiệm Chuyên đề toán 9 Chuyên đề 2-1: Biểu thức chứa chữ có đáp án

Điều kiện: $\left\{\begin{array}{l}x\ge 0\\ x-9\ne 0\\ \sqrt{x}+3\ne 0\end{array}\right.\iff \left\{\begin{array}{l}x\ge 0\\ x\ne 9\end{array}\right.$ 

Với $x\ge 0,x\ne 9$ ta có:

$$\begin{gathered} P=\left(\frac{\sqrt{x}+1}{(\sqrt{x}\text{ +3)(}\sqrt{x}\text{ -3)}}-\frac{1}{\sqrt{x}+3}\right)\left(\sqrt{x}-3\right)=\left[\frac{\sqrt{x}+1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\left(\sqrt{x}-3\right) \\ =\frac{\sqrt{x}+1-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\left(\sqrt{x}-3\right)=\frac{4}{\sqrt{x}+3} \end{gathered}$$

Vậy $P=\frac{4}{\sqrt{x}+3}$ với $x\ge 0,x\ne 9$.

Để Q có nghĩa, điều kiện là: $\left\{\begin{array}{l}x\ge 0\\ x-\sqrt{x}-2\ne 0\\ \sqrt{x}-2\ne 0\\ \sqrt{x}+1\ne 0\end{array}\right.\iff \left\{\begin{array}{l}x\ge 0\\ \sqrt{x}\ne 2\end{array}\right.\iff \left\{\begin{array}{l}x\ge 0\\ x\ne 4\end{array}\right.$ 

Với điều kiện trên ta có: 

$$\begin{gathered} Q=\frac{\sqrt{x}-1}{x-\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{2\sqrt{x}-1}{\sqrt{x}-2}=\frac{\sqrt{x}-11}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{2\sqrt{x}-1}{\sqrt{x}-2} \\ =\frac{\sqrt{x}-11-\sqrt{x}\left(\sqrt{x}-2\right)+\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-11-\left(x-2\sqrt{x}\right)+\left(2x+\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)} \\ =\frac{x+4\sqrt{x}-12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+6}{\sqrt{x}+1} \end{gathered}$$

Thay x = 9 vào A ta được: $A=\frac{\sqrt{9}+1}{\sqrt{9}-1}=\frac{3+1}{3-1}=\frac{4}{2}=2$ 

Với $x\ge 0,x\ne 4$ ta có:

$$\begin{gathered} A=\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}=\frac{\left(2x-4\sqrt{x}\right)+\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=\frac{2\sqrt{x}\left(\sqrt{x}-2\right)+\left(\sqrt{x}-2\right)}{\sqrt{x}-2} \\ =\frac{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=2\sqrt{x}+1 \end{gathered}$$

Khi $x=4-2\sqrt{3}={\left(\sqrt{3}\right)}^2-2.\sqrt{3}.1+1^2={\left(\sqrt{3}-1\right)}^2$ 

$\Rightarrow \sqrt{x}=\sqrt{3}-1$, thay vào A ta được:

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

Vậy $x=4-2\sqrt{3}$ thì $A=2\sqrt{3}-1$.

Ta thấy $x=4-2\sqrt{3}$ có thể rút gọn bằng cách đưa về bình phương của một hiệu. Do vậy, ta cần rút gọn x trước khi thay vào biểu thức A.

a) Với $x\ge 0,x\ne 1$ ta có:

Vậy $B=\frac{2\sqrt{x}}{x-1}$ với $x\ge 0,x\ne 1$.

b) Ta có: $x=\sqrt{17+12\sqrt{2}}=\sqrt{9+\mathrm{2.3.2}\sqrt{2}+8}=\sqrt{{\left(3+2\sqrt{2}\right)}^2}=3+2\sqrt{2}$ (thỏa mãn điều kiện $x\ge 0,x\ne 1$).

$\sqrt{x}=\sqrt{3+2\sqrt{2}}=\sqrt{2+2.\sqrt{2}.1+1}=\sqrt{{\left(\sqrt{2}+1\right)}^2}=\sqrt{2}+1$ 

Thay $x=\sqrt{2}+1$ vào B ta được: $B=\frac{2\left(\sqrt{2}+1\right)}{3+2\sqrt{2}-1}=\frac{2\left(\sqrt{2}+1\right)}{2\left(1+\sqrt{2}\right)}=1$ 

Với $x\ne 1;x\ge 0$ ta có:

$$\begin{gathered} C=\frac{{\left(\sqrt{x}\right)}^3+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{x-1}{\sqrt{x}+1}=\frac{{\left(\sqrt{x}\right)}^3+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1} \\ =\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\frac{x-\sqrt{x}+1}{\sqrt{x}-1}-\left(\sqrt{x}-1\right) \\ =\frac{x-\sqrt{x}+1-{\left(\sqrt{x}-1\right)}^2}{\sqrt{x}-1}=\frac{x-\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}=\frac{\sqrt{x}}{\sqrt{x}-1} \end{gathered}$$

Vậy $C=\frac{\sqrt{x}}{\sqrt{x}-1}$ với $x\ne 1;x\ge 0$.

Suy ra $C-1=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\frac{1}{\sqrt{x}-1}$

Ta có $x=2020+2\sqrt{2019}$ (thỏa mãn điều kiện $x\ne 1,x\ge 0$).

Có $x=2019+2\sqrt{2019}+1={\left(\sqrt{2019}\right)}^2+2.\sqrt{2019}.1+1^2={\left(\sqrt{2019}+1\right)}^2\Rightarrow \sqrt{x}=\sqrt{2019}+1$

Thay vào biểu thức C - 1 ta được : $C-1=\frac{1}{\sqrt{2019}+1-1}=\frac{1}{\sqrt{2019}}$

Vậy $C-1=\frac{1}{\sqrt{2019}}$ khi $x=2020+2\sqrt{2019}$.

Với x > 0 và $x\ne 1$ta có:

$$\begin{gathered} VT=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)} \\ =\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}} \end{gathered}$$

Vậy với với x > 0 và $x\ne 1$ thì $\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}}$.

Với x > 0 và $x\ne 4$ta có:

$$\begin{gathered} P=\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{2}{x-4}\right).\left(\sqrt{x}-1+\frac{\sqrt{x}-4}{\sqrt{x}}\right) \\ =\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{x-4}-\frac{2}{x-4}\right].\left[\frac{\left(\sqrt{x}-1\right)\sqrt{x}}{\sqrt{x}}+\frac{\sqrt{x}-4}{\sqrt{x}}\right] \\ =\frac{x+2\sqrt{x}+\sqrt{x}+2-2}{x-4}.\frac{x-\sqrt{x}+\sqrt{x}-4}{\sqrt{x}}=\frac{x+3\sqrt{x}}{x-4}.\frac{x-4}{\sqrt{x}}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}}=\sqrt{x}+3 \end{gathered}$$

Vậy $P=\sqrt{x}+3$ với mọi $x>0,x\ne 4$.

a) Với $a>0,a\ne 1$ ta có:

$$\begin{gathered} P=\left[\frac{1}{a-1}+\frac{3\sqrt{a}+5}{a\left(\sqrt{a}-1\right)-\left(\sqrt{a}-1\right)}\right].\left[\frac{(a+2\sqrt{a}+1)-4\sqrt{a}}{4\sqrt{a}}\right]=\left[\frac{1}{a-1}+\frac{3\sqrt{a}+5}{\left(a-1\right)\left(\sqrt{a}-1\right)}\right].\frac{a-2\sqrt{a}+1}{4\sqrt{a}} \\ =\frac{\sqrt{a}-1+3\sqrt{a}+5}{\left(a-1\right)\left(\sqrt{a}-1\right)}.\frac{{\left(\sqrt{a}-1\right)}^2}{4\sqrt{a}}=\frac{4\sqrt{a}+4}{\left(a-1\right)\left(\sqrt{a}-1\right)}.\frac{{\left(\sqrt{a}-1\right)}^2}{4\sqrt{a}}=\frac{4\sqrt{a}+4}{a-1}.\frac{\sqrt{a}-1}{4\sqrt{a}} \\ =\frac{4\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}-1}{4\sqrt{a}}=\frac{1}{\sqrt{a}} \end{gathered}$$

b) Ta có: $Q=\left(a-\sqrt{a}+1\right)P=\frac{a-\sqrt{a}+1}{\sqrt{a}}$

Xét $Q-1=\frac{a-\sqrt{a}+1}{\sqrt{a}}-1=\frac{a-\sqrt{a}+1-\sqrt{a}}{\sqrt{a}}=\frac{a-2\sqrt{a}+1}{\sqrt{a}}=\frac{{\left(\sqrt{a}-1\right)}^2}{\sqrt{a}}$ 

Vì ${\left(\sqrt{a}-1\right)}^2>0$ và $\sqrt{a}>0,\forall a>0,a\ne 1$ nên $Q-1>0\Rightarrow Q>1$ 

Với $x\ge 0,x\ne 1$ ta có:

$$\begin{gathered} B=\left(\frac{\sqrt{x}-1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+1}\right):\frac{3}{\sqrt{x}+1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{3} \\ =\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{3}=\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{3}=\frac{1}{\sqrt{x}-2} \\ \Rightarrow P=\frac{18}{A.B}=\frac{18\left(\sqrt{x}-1\right)}{\sqrt{x}+2}=18-\frac{54}{\sqrt{x}+2} \end{gathered}$$

Vì $x\ge 0\Rightarrow \sqrt{x}+2\ge 2\Rightarrow \frac{54}{\sqrt{x}+2}\le \frac{54}{2}=27$ nên $P=18-\frac{54}{\sqrt{x}+2}\ge 18-27=-9$

Hay $P\ge -9,\forall x\ge 0$ 

Đẳng thức xảy ra khi và chỉ khi x = 0 

Với $x\ge 0,x\ne 1$ ta có:

$$\begin{gathered} A=\left(\frac{1}{1-\sqrt{x}}+\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}\right):\frac{\sqrt{x}-1}{3} \\ =\frac{-\left(x+\sqrt{x}+1\right)+x+2+\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{3}{\sqrt{x}-1} \\ =\frac{3{\left(\sqrt{x}-1\right)}^2}{{\left(\sqrt{x}-1\right)}^2\left(x+\sqrt{x}+1\right)}=\frac{3}{x+\sqrt{x}+1} \end{gathered}$$

Vậy $A=\frac{3}{x+\sqrt{x}+1}$ với $x\ge 0,x\ne 1$

A đạt giá trị lớn nhất $\iff x+\sqrt{x}+1$ đạt giá trị nhỏ nhất

Vì $x\ge 0$ nên $x+\sqrt{x}+1\ge 1\Rightarrow A=\frac{3}{x+\sqrt{x}+1}\le 3$ 

Đẳng thức xảy ra <=> x = 0. Vậy maxA = 3 khi x = 0.

Với $x>0,x\ne 4$ ta có:

$$\begin{gathered} Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{x-4}=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{x-4} \\ =\frac{x+2\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x}-2} \end{gathered}$$

$\Rightarrow \frac{P}{Q}=\frac{x+3}{\sqrt{x}}=\sqrt{x}+\frac{3}{\sqrt{x}}\ge 2\sqrt{3}$  (Do bất đẳng thức Cô-si).

Đẳng thức xảy ra khi $\sqrt{x}=\frac{3}{\sqrt{x}}\iff x=3$

Vậy giá trị nhỏ nhất của $\frac{P}{Q}$ là $2\sqrt{3}$ khi x = 3.

Điều kiện: $x\ge 0,x\ne 9$.

a) Với $x\ge 0,x\ne 9$ ta có:

$$\begin{gathered} P=\frac{3\sqrt{x}+2}{\sqrt{x}+1}-\frac{2\sqrt{x}-3}{3-\sqrt{x}}-\frac{3\left(3\sqrt{x}-5\right)}{(\sqrt{x}+1)(\sqrt{x}-3)} \\ =\frac{\left(3\sqrt{x}+2\right)\left(3-\sqrt{x}\right)+\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)-3\left(3\sqrt{x}-5\right)}{(\sqrt{x}+1)(\sqrt{x}-3)} \\ =\frac{3x-9\sqrt{x}+2\sqrt{x}-6+2x+2\sqrt{x}-3\sqrt{x}-3-9\sqrt{x}+15}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)} \\ =\frac{5x-17\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{5x-15\sqrt{x}-2\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}-3\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{5\sqrt{x}-2}{\sqrt{x}+1} \end{gathered}$$

b) Ta có $P=\frac{5\sqrt{x}-2}{\sqrt{x}+1}=\frac{5\sqrt{x}+5-7}{\sqrt{x}+1}=5-\frac{7}{\sqrt{x}+1}$      

Vì $\frac{7}{\sqrt{x}+1}>0$ nên P có giá trị nhỏ nhất $\iff \frac{7}{\sqrt{x}+1}$ lớn nhất $\iff \sqrt{x}+1$ nhỏ nhất <=> x = 0 

Khi đó minP = 5 - 7 = -2 

Với $x\ge 0,x\ne \frac{1}{4},x\ne 1,x\ne 4$ ta có:

$$\begin{gathered} P=\left(\frac{x-4}{x-3\sqrt{x}+2}+1\right):\frac{1}{2x-3\sqrt{x}+1}=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+1\right].\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right) \\ =\left(\frac{\sqrt{x}+2}{\sqrt{x}-1}+1\right)\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)=\left(\sqrt{x}+2+\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)=4x-1 \end{gathered}$$

Vậy P = 4x - 1 với $x\ge 0,x\ne \frac{1}{4},x\ne 1,x\ne 4$.

b) Xét $T=P+\frac{10}{x}=4x-1+\frac{10}{x}=\frac{2x}{5}+\frac{10}{x}+\frac{10x}{5}-1$

Áp dung bất đẳng thức Cô-si ta có: $\frac{2x}{5}+\frac{10}{x}\ge 2\sqrt{\frac{2x}{5}.\frac{10}{x}}=4$ 

Đẳng thức xảy ra $\iff \frac{2x}{5}=\frac{10}{x}\iff x=5$ (do $x\ge 0$).

Lại có: $\frac{18x}{5}\ge 18$ (vì $x\ge 5$ ) nên $T\ge 4+18-1=21$ 

a) Điều kiện: $x>0,x\ne 1$.

$$\begin{gathered} A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\frac{x-2\sqrt{x}+1}{x-1} \\ =\left[\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\frac{{\left(\sqrt{x}-1\right)}^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)} \\ =\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1} \end{gathered}$$

Vậy $A=\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}$ với $x>0,x\ne 1$.

b) $\left|A\right|>A\iff A<0\iff \frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}<0\iff \sqrt{x}-1<0\iff x<1$

Kết hợp với điều kiện ta được 0 < x < 1 thì $\left|A\right|>A$.

c) Ta có: $A=\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=2+\frac{4}{\sqrt{x}-1}$ 

Để A nguyên thì $\sqrt{x}-1\in U\left(4\right)$ hay $\sqrt{x}-1\in \left\{-1;1;-2;2;-4;4\right\}$

Ta có bảng sau:

Vậy $x\in \left\{4;9;25\right\}$.

Điều kiện: $x>0,x\ne 4,x\ne 9$.

a) Với $x>0,x\ne 4,x\ne 9$ ta có:

$$\begin{gathered} B=\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{5\sqrt{x}+2}{(2+\sqrt{x})(2-\sqrt{x})}\right):\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{{\left(\sqrt{x}+2\right)}^2} \\ =\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2\sqrt{x}\left(\sqrt{x}-2\right)-\left(5\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{{\left(\sqrt{x}+2\right)}^2}{\sqrt{x}\left(3-\sqrt{x}\right)} \\ =\frac{x+2\sqrt{x}+\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{{\left(\sqrt{x}+2\right)}^2}{\sqrt{x}\left(3-\sqrt{x}\right)} \\ =\frac{-x+2\sqrt{x}}{\sqrt{x}-2}.\frac{\sqrt{x}+2}{\sqrt{x}\left(3-\sqrt{x}\right)}=\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-2}.\frac{\sqrt{x}+2}{\sqrt{x}\left(3-\sqrt{x}\right)}=\frac{\sqrt{x}+2}{\sqrt{x}-3} \end{gathered}$$

Vậy $B=\frac{\sqrt{x}+2}{\sqrt{x}-3}$ với $x>0,x\ne 4,x\ne 9$.

c) $B<0\iff \frac{\sqrt{x}+2}{\sqrt{x}-3}<0\iff \sqrt{x}-3<0$ (vì $\sqrt{x}+2>0$)

a) Với x > 0, $x\ne 4$ ta có:

$$\begin{gathered} P=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right) \\ =\left(\frac{x}{\sqrt{x}\left(x-4\right)}-\frac{6}{3\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right) \\ =\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+2}{6} \\ =\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+2}{6}=\frac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+2}{6}=\frac{-1}{\sqrt{x}-2} \end{gathered}$$

Vậy $P=\frac{-1}{\sqrt{x}-2}$ với x > 0, $x\ne 4$.

b) Với $x>0,x\ne 4$ ta có $Q=\left(-\sqrt{x}-1\right).P=\left(-\sqrt{x}-1\right).\frac{-1}{\sqrt{x}-2}=\frac{\sqrt{x}+1}{\sqrt{x}-2}=1+\frac{3}{\sqrt{x}-2}$

Để Q nguyên thì $\frac{3}{\sqrt{x}-2}\in \mathbb{Z} \iff \sqrt{x}-2\in Ư\left(4\right)$ hay $\sqrt{x}-2\in \left\{-1;1;-3;3\right\}$

Ta có bảng:

b) ta có: 

$$\begin{gathered} \frac{A}{B}=m+1\iff \frac{\sqrt{x}+3}{\sqrt{x}-4}:\frac{\sqrt{x}}{\sqrt{x}-4}=m+1\iff \frac{\sqrt{x}+3}{\sqrt{x}}=m+1 \\ \iff \frac{\sqrt{x}+3-\sqrt{x}\left(m+1\right)}{\sqrt{x}}=0\iff \frac{-m\sqrt{x}+3}{\sqrt{x}}=0 \end{gathered}$$ 

Để phương trình $\frac{A}{B}=m+1$ có nghiệm thì phương trình $\frac{-m\sqrt{x}+3}{\sqrt{x}}=0$ có nghiệm, tức là:

$\frac{-m\sqrt{x}+3}{\sqrt{x}}=0$ có nghiệm $\iff \left\{\begin{array}{l}x>0\\ -m\sqrt{x}+3=0\\ x\ne 16\end{array}\right.\iff \left\{\begin{array}{l}x>0\\ \sqrt{x}=\frac{3}{m}\\ x\ne 16\end{array}\right.\iff \left\{\begin{array}{l}\frac{3}{m}>0\\ \frac{3}{m}\ne 4\end{array}\right.\iff \left\{\begin{array}{l}m>0\\ m\ne \frac{3}{4}\end{array}\right.$ 

$B=\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}}\right):\frac{\sqrt{x}+1}{\sqrt{x}}=(\sqrt{x}-\frac{1}{\sqrt{x}}):\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{x-1}{\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1$

a) Thay x = 25 (thỏa mãn điều kiện) vào A ta được: $A=\frac{25+\sqrt{25}+1}{\sqrt{25}-4}=31$.