Dạng 5. Bài toán chứa ẩn (ẩn x) dưới dấu căn và những ý toán phụ.

$Q=2\iff =\frac{\sqrt{x}+2}{\sqrt{x}-3}=2$

$\iff \sqrt{x}+2=2\sqrt{x}-6$

$\iff -\sqrt{x}=-8\iff \sqrt{x}=8\iff x=64$ (Thỏa mãn ĐKXĐ).

$Q<0\iff \frac{\sqrt{x}+2}{\sqrt{x}-3}<0$

Để $B\in Z\iff \frac{11}{a-9}\in Z\iff 11⋮(a-9)\iff (a-9)\in U\left(11\right)$

$U\left(11\right)=\left\{1;11;-1;-11\right\}$

Khi đó ta có bảng giá trị

$$\begin{gathered} \text{P}=\frac{\left(x-9\right)+\left(4-x\right)+\left(9-x\right)}{\left(2-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\frac{x-9-3\sqrt{x}+9}{x-9} \\ =\frac{4-x}{\left(2-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}=\frac{2+\sqrt{x}}{\sqrt{x}} \end{gathered}$$

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

Nên $\text{P}=\frac{2+\sqrt{2}}{\sqrt{2}}=\sqrt{2}+1$

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

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

$\frac{A}{B}\ge \frac{x}{4}+5\iff \frac{\sqrt{x}+4}{\sqrt{x}-1}:\frac{1}{\sqrt{x}-1}\ge \frac{x}{4}+5$

$\iff 4(\sqrt{x}+4)\ge x+20\iff x-4\sqrt{x}+4\le 0\iff {\left(\sqrt{x}-2\right)}^2\le 0\iff \sqrt{x}-2=0\iff x=4$

$A=\frac{\sqrt{x}+2}{x+\sqrt{x}+1}.$

Có $x=\sqrt{2022+4\sqrt{2018}}-\sqrt{2022-4\sqrt{2018}}$

$=\sqrt{{\left(\sqrt{2018}+2\right)}^2}-\sqrt{{\left(\sqrt{2018}-2\right)}^2}$

$=\left|\sqrt{2018}+2\right|-\left|\sqrt{2018}-2\right|=\sqrt{2018}+2-\sqrt{2018}+2=4$ thỏa mãn điều kiện  x > 0 và $x\ne 1$

Với $a>0;a\ne 1$, ta có:

$B=\left[\frac{6}{a-1}+\frac{10-2\sqrt{a}}{(a-1)(\sqrt{a}-1)}\right].\frac{{(\sqrt{a}-1)}^2}{4\sqrt{a}}$

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

$B=\frac{1}{\sqrt{a}}.$

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

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

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

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

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

Ta có: $B=\left[\frac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+3}{\sqrt{x}-1}\right].\frac{x-1}{2x+\sqrt{x}-1}$

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

$=\frac{2\sqrt{x}+3}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{2\sqrt{x}-1}=\frac{2\sqrt{x}+3}{2\sqrt{x}-1}$

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

$V=\frac{1}{3}\iff \frac{2}{\sqrt{x}-2}=\frac{1}{3}\iff \sqrt{x}-2=6\iff x=64$(thỏa mãn)

Ta có:

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

Với $x\ge 0,x\ne 1,x\ne 4$, ta có:

$$\begin{gathered} Q=\frac{\left(x+27\right).P}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)} =\frac{x+27}{\sqrt{x}+3} =\frac{x-9+36}{\sqrt{x}+3} \\ =\sqrt{x}-3+\frac{36}{\sqrt{x}+3} =-6+\left(\sqrt{x}+3\right)+\frac{36}{\sqrt{x}+3}\ge -6+12=6 (co-si) \end{gathered}$$

Dấu "=" xảy ra khi $\sqrt{x}+3=\frac{36}{\sqrt{x}+3}$ $\iff {\left(\sqrt{x}+3\right)}^2=36$ $\iff x=9$