Chuyên đề 1: Căn bậc 2 có đáp án

$A=\sqrt{16}-\sqrt{25}+\sqrt{4}=4-5+2=1<\sqrt{2}$

Vậy $A<\sqrt{2}$

$2\sqrt{9}-3\sqrt{4}=2.3-3.2=0$

$\sqrt{\frac{28{(a-2)}^2}{7}}\text{\hspace{0.17em}=}\sqrt{4{(a-2)}^2}$

$=2.\left|\right(a-2\left)\right|=2 (a-2). Do a>2 nên a-2>0 \Rightarrow |a-2|=a-2$

$A=\sqrt{20}-\sqrt{45}+3\sqrt{80}$$=2\sqrt{5}-3\sqrt{5}+12\sqrt{5}=11\sqrt{5}$

$T=\sqrt{4}+\sqrt{25}-\sqrt{9}=2+5-3=4$

Chọn D

Thay x = 7 (thỏa mãn) vào biểu thức $\frac{4}{\sqrt{x+2}-1}$ ta tính được biểu thức có giá trị bằng

$\frac{4}{\sqrt{7+2}-1}=\frac{4}{3-1}=2$

$$\begin{gathered} A=2\sqrt{48}+3\sqrt{75}-2\sqrt{108} \\ A=2\sqrt{4^2.3}+3\sqrt{5^2.3}-2\sqrt{6^2.3} \\ A=\mathrm{2.4.}\sqrt{3}+3.5\sqrt{3}-2.6\sqrt{3} \\ A=8\sqrt{3}+15\sqrt{3}-12\sqrt{3} \\ A=(8+15-12)\sqrt{3}=11\sqrt{3} \end{gathered}$$

Vậy $A=11\sqrt{3}.$

$$\begin{gathered} B=\sqrt{19+8\sqrt{3}}+\sqrt{19-8\sqrt{3}} \\ B=\sqrt{4^2+\mathrm{2.4.}\sqrt{3}+{\left(\sqrt{3}\right)}^2}+\sqrt{4^2-\mathrm{2.4.}\sqrt{3}+{\left(\sqrt{3}\right)}^2} \\ B=\sqrt{{(4+\sqrt{3})}^2}+\sqrt{{(4-\sqrt{3})}^2} \\ B=|4+\sqrt{3}|+|4-\sqrt{3}| \\ B=4+\sqrt{3}+4-{\sqrt{3}}^{}(4+\sqrt{3}>0;4-\sqrt{3}>0) \\ B=8 \end{gathered}$$

Vậy B = 8.

$$\begin{gathered} A=3\sqrt{49}-\sqrt{25} \\ A=3\sqrt{7^2}-\sqrt{5^2} \\ A=3.7-5 \\ A=21-5 \\ A=16 \end{gathered}$$

$$\begin{gathered} B=\sqrt{{(3-2\sqrt{5})}^2}-\sqrt{20} \\ B=\left|3-2\sqrt{5}\right|-\sqrt{2^2.5} \\ B=-(3-2\sqrt{5})-2\sqrt{5} \\ B=-3+2\sqrt{5}-2\sqrt{5} \\ B=-3 \end{gathered}$$

$A=\sqrt{45}-2\sqrt{20}=\sqrt{3^2.5}-2\sqrt{2^2.5}=3\sqrt{5}-2.2\sqrt{5}=-\sqrt{5}$

$$\begin{gathered} B=\frac{3\sqrt{5}-\sqrt{27}}{\sqrt{3}-\sqrt{5}}-\sqrt{{\left(3-\sqrt{12}\right)}^2}=\frac{3\sqrt{5}-3\sqrt{3}}{\sqrt{3}-\sqrt{5}}-\left|3-\sqrt{12}\right| \\ =\frac{3\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{3}-\sqrt{5}}-\left(-3+\sqrt{12}\right) \end{gathered}$$

(do $3^2<12\Rightarrow 3<\sqrt{12}$)

$=-3+3-\sqrt{12}=-\sqrt{12}=-2\sqrt{3}$

$$\begin{gathered} A=\sqrt{27}-\sqrt{12} \\ A=3\sqrt{3}-2\sqrt{3} \\ A=\sqrt{3} \end{gathered}$$

Chọn B

$P=2\sqrt{27}+\sqrt{300}-3\sqrt{75}=6\sqrt{3}+10\sqrt{3}-15\sqrt{3}=\sqrt{3}$

Chọn B

Biểu thức $\sqrt{2x-4}$ có nghĩa khi và chỉ khi: $2x-4\ge 0\iff x\ge 2$

$A=\sqrt{32}-\sqrt{6}.\sqrt{3}+\frac{\sqrt{22}}{\sqrt{11}}=4\sqrt{2}-\sqrt{2}.\sqrt{3}.\sqrt{3}+\sqrt{\frac{22}{11}}$

$=4\sqrt{2}-3\sqrt{2}+\sqrt{2}$

$=2\sqrt{2}$

$A={\left(\sqrt{5}\right)}^2-3^2+6=5-9+6=2$

$P=\sqrt{5}(\sqrt{5}+2)-\sqrt{20}=\sqrt{5}.\sqrt{5}+2\sqrt{5}-2\sqrt{5}=5$

Vậy P = 5.

$$\begin{gathered} A=\frac{4+\sqrt{8}+\sqrt{2}-\sqrt{3}-\sqrt{6}}{2+\sqrt{2}-\sqrt{3}} \\ =\frac{4+2\sqrt{2}+\sqrt{2}-\sqrt{3}-\sqrt{2}\sqrt{3}}{2+\sqrt{2}-\sqrt{3}} \\ =\frac{4+3\sqrt{2}-\sqrt{3}-\sqrt{2}\sqrt{3}}{2+\sqrt{2}-\sqrt{3}} \\ =\frac{\left(2+\sqrt{2}-\sqrt{3}\right)+\left(2\sqrt{2}+2-\sqrt{2}.\sqrt{3}\right)}{2+\sqrt{2}-\sqrt{3}} \\ =\frac{\left(2+\sqrt{2}-\sqrt{3}\right)+\sqrt{2}\left(2+\sqrt{2}-\sqrt{3}\right)}{2+\sqrt{2}-\sqrt{3}} \\ =\frac{\left(2+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}\right)}{2+\sqrt{2}-\sqrt{3}} \\ =1+\sqrt{2} \end{gathered}$$

Vậy A = $=1+\sqrt{2}$

$3\sqrt{4}+2\sqrt{25}-4\sqrt{9}=3.2+2.5-4.3=4$

$3\sqrt{3}+5\sqrt{12}-2\sqrt{27}=3\sqrt{3}+5.2\sqrt{3}-2.3\sqrt{3}=3\sqrt{3}+10\sqrt{3}-6\sqrt{3}=7\sqrt{3}$

$K=\sqrt{9}+\sqrt{45}-3\sqrt{5}=3+3\sqrt{5}-3\sqrt{5}=3$

$Q=\frac{x-4}{\sqrt{x}+2}+\frac{x+2\sqrt{x}}{\sqrt{x}}=\frac{\left(\sqrt{x}+2\right).\left(\sqrt{x}-2\right)}{\sqrt{x}+2}+\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}=\sqrt{x}-2+\sqrt{x}+2=2\sqrt{x}$

$\sqrt{4}+3=2+3=5$

$\sqrt{5}+\sqrt{{\left(6-\sqrt{5}\right)}^2}=\sqrt{5}+\left|6-\sqrt{5}\right|=\sqrt{5}+6-\sqrt{5}=6$

$\sqrt{27}+4\sqrt{3}-\sqrt{3}=3\sqrt{3}+8\sqrt{3}-\sqrt{3}=10\sqrt{3}$

$A=\sqrt{16}-\sqrt{4}=4-2=2$

$B=\sqrt{5}\left(\sqrt{5}-3\right)+3\sqrt{5}=5-3\sqrt{5}+3\sqrt{5}=5$

$C=\sqrt{{\left(\sqrt{2}-5\right)}^2}+\sqrt{2}=\left|\sqrt{2}-5\right|+\sqrt{2}=-(\sqrt{2}-5)+\sqrt{2}=-\sqrt{2}+5+\sqrt{2}=5$

$A=\left(\sqrt{12}-2\sqrt{5}\right)\sqrt{3}+\sqrt{60}=\sqrt{36}-2\sqrt{15}+2\sqrt{15}=\sqrt{36}=6$

$P=\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{3}{\sqrt{x}-3}+\frac{6\sqrt{x}}{x-9}$

$\begin{array}{l}=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\frac{x-3\sqrt{x}-3\sqrt{x}-9+6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\frac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=1\end{array}$

Vậy P = 1.

$\begin{array}{l}A=\sqrt{12}+\sqrt{{\left(\sqrt{2}-1\right)}^2}-\frac{1}{\sqrt{3}-\sqrt{2}}\\ A=2\sqrt{3}+\left|\sqrt{2}-1\right|-\frac{\sqrt{3}+\sqrt{2}}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}\\ =2\sqrt{3}+\sqrt{2}-1-\sqrt{3}-\sqrt{2}\\ =\sqrt{3}-1\end{array}$

$\begin{array}{l}B=\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}\\ B=\frac{1}{\sqrt{x}(\sqrt{x}+1)}+\frac{2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{1}{\sqrt{x}(\sqrt{x}-1)}\\ B=\frac{\sqrt{x}-1+2x-\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+1)}\\ B=\frac{2x-2}{\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2}{\sqrt{x}}\end{array}$

$B=8\iff \frac{2}{\sqrt{x}}=8\iff \sqrt{x}=\frac{1}{4}\iff x=\frac{1}{6}\left(TMĐK\right)$

a) $A=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{1}{\sqrt{x}-2}$

$$\begin{gathered} =\frac{x-4-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)} \\ =\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)} \\ =\frac{\sqrt{x}-4}{\sqrt{x}-2} \end{gathered}$$

b) $x=6+4\sqrt{2}={\left(2+\sqrt{2}\right)}^2$ (TMĐK)

$\sqrt{x}=2+\sqrt{2}$ thay vào A ta được $A=\frac{\left(2+\sqrt{2}\right)-4}{\left(2+\sqrt{2}\right)-2}=\frac{2-\sqrt{2}}{2}=1-\sqrt{2}$

Vậy với $x=6+4\sqrt{2}$ thì $A=1-\sqrt{2}$

Ta có A có giá trị dương $\iff A>0\iff x-1>0\iff x>1$

Vậy x > 1 thì A có giá trị dương

$$\begin{gathered} B=2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5} \\ =2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5}=2.2\sqrt{5}-3.3\sqrt{5}+4.4\sqrt{5} \\ =4\sqrt{5}-9\sqrt{5}+16\sqrt{5}=11\sqrt{5} \end{gathered}$$

Vậy B = $11\sqrt{5}$

ĐKXĐ: $a\ge 0$; $a\ne 1$

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

Vậy với $a\ge 0$; $a\ne 1$ thì B = 1.

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

$x=2\Rightarrow A=2\sqrt{2}-1$

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

$$\begin{gathered} A=\sqrt{2\sqrt{5}+6}-\sqrt{{(\sqrt{5}-1)}^2}+2018 \\ \begin{array}{l}=\sqrt{{\left(\sqrt{5}+1\right)}^2}-\sqrt{{\left(\sqrt{5}-1\right)}^2}+2018\\ =\sqrt{5}+1-\sqrt{5}+1+2018\\ =2020\end{array} \end{gathered}$$

Vậy A là một số nguyên

$$\begin{gathered} P=\frac{a-1}{\sqrt{b}-1}\sqrt{\frac{b-2\sqrt{b}+1}{a^2-2a+1}} \\ \begin{array}{l}=\frac{a-1}{\sqrt{b}-1}\sqrt{\frac{{\left(\sqrt{b}-1\right)}^2}{{\left(a-1\right)}^2}}\\ =\frac{a-1}{\sqrt{b}-1}.\frac{\left|\sqrt{b}-1\right|}{\left|a-1\right|}\\ =\frac{a-1}{\sqrt{b}-1}.\frac{\sqrt{b}-1}{1-a}\\ =-1\end{array} \end{gathered}$$

(do a < 1 và b > 1)

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

b) $2019A=\frac{2019\left(2\sqrt{x}+2-3\right)}{\sqrt{x}+1}=4038-\frac{6057}{\sqrt{x}+1}$

2019A là số nguyên khi và chỉ khi $\sqrt{x}+1$ là ước nguyên dương của 6057 gồm: $1;3;9;673,2019;6057$.

+) $\sqrt{x}+1=1\iff x=0$, thỏa mãn.

+) $\sqrt{x}+1=3\iff x=4$, thỏa mãn.

+) $\sqrt{x}+1=9\iff x=64$, thỏa mãn.

+) $\sqrt{x}+1=673\iff x=451584$, thỏa mãn.

+) $\sqrt{x}+1=2019\iff x=4072324$, thỏa mãn.

+) $\sqrt{x}+1=6057\iff x=36675136$, thỏa mãn.

a) $P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-\sqrt{x}}\right):\frac{\sqrt{x}+1}{3}$

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

b) $P=1\iff \frac{3}{\sqrt{x}-1}=1$

$$\begin{gathered} \iff \sqrt{x}-1=3 \\ \iff \sqrt{x}=4 \\ \iff x=16 \end{gathered}$$

Vậy x = 16 thì P = 1.

Khi x = 5, ta có 

$$\begin{gathered} A=\sqrt{5+2\sqrt{5-1}}+\sqrt{5-2\sqrt{5-1}} \\ =\sqrt{5}+2\sqrt{4}+\sqrt{5-2\sqrt{4}}=\sqrt{5+2\cdot 2}+\sqrt{5-2\cdot 2}=\sqrt{9}+\sqrt{1}=3+1=4 \end{gathered}$$

Vậy khi x = 5 thì A = 4.

Với $1\le x\le 2$, ta có

$$\begin{gathered} A=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}} \\ =\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1} \\ =\sqrt{{(\sqrt{x-1}+1)}^2}+\sqrt{{(\sqrt{x-1}-1)}^2} \\ =|\sqrt{x-1}+1|+|\sqrt{x-1}-1| \\ =\sqrt{x-1}+1+1-\sqrt{x-1}\text{\hspace{1em} (}1\le x\le 2\Rightarrow 0\le \sqrt{x-1}\le 1\Rightarrow \sqrt{x-1}-1\le 0) \\ =2 \end{gathered}$$

Vậy khi $1\le x\le 2$ thì A = 2.

Với $x>0;x\ne 4$ ta có

$A=\left[\frac{2x-4\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}$

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

Kết luận A = $=\frac{1}{\sqrt{x}+2}$

$$\begin{gathered} A=\frac{4}{\sqrt{5}-1}-3\sqrt{45}+\sqrt{{\left(\sqrt{5}-1\right)}^2}=\frac{4\left(\sqrt{5}+1\right)}{5-1}-9\sqrt{5}+\left|\sqrt{5}-1\right| \\ =\sqrt{5}+1-9\sqrt{5}+\sqrt{5}-1=-7\sqrt{5} \end{gathered}$$

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

$B>\frac{1}{2}\iff \frac{2}{3-\sqrt{x}}>\frac{1}{2}\iff \frac{2}{3-\sqrt{x}}-\frac{1}{2}>0\iff \frac{4-\left(3-\sqrt{x}\right)}{2\left(3-\sqrt{x}\right)}>0$

$\iff \frac{1+\sqrt{x}}{2\left(3-\sqrt{x}\right)}>0;\left(*\right)$

Vì $1+\sqrt{x}>0$ nên $\left(*\right)\iff 3-\sqrt{x}>0\iff \sqrt{x}<3\iff 0<x<9$

Vì $x\in \mathbb{Z}\Rightarrow x\in \left\{1;2;3;4;5;6;7;8\right\}$

a) Với x = 9

Thay vào A ta có : $A=\frac{4\left(\sqrt{x}+1\right)}{25-x}=\frac{4\left(\sqrt{9}+1\right)}{25-9}=\frac{4.\left(3+1\right)}{16}=1$

b)

Với $x\ge 0;x\ne 25$ ta có         

$$\begin{gathered} B=\left(\frac{15-\sqrt{x}}{x-25}+\frac{2}{\sqrt{x}+5}\right):\frac{\sqrt{x}+1}{\sqrt{x}-5} \\ B=\left[\frac{15-\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\frac{2}{\sqrt{x}+5}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-5} \\ B=\frac{15-\sqrt{x}+2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-5} \\ B=\frac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-5} \\ B=\frac{\sqrt{x}+5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot \frac{\sqrt{x}-5}{\sqrt{x}+1} \\ B=\frac{1}{\sqrt{x}+1} \end{gathered}$$