Cho hai biểu thức = căn x+1/x-4 và

1) Khi x = 25 (TMĐK) ta có:

$A=\frac{\sqrt{25}+1}{25-4}=\frac{5+1}{21}=\frac{6}{21}=\frac{2}{7}$.

Vậy giá trị của A khi x = 25 là $A=\frac{2}{7}$.

2) $B=\frac{18-\sqrt{x}}{x-4}+\frac{4}{2-\sqrt{x}}+\frac{\sqrt{x}+3}{\sqrt{x}+2}$

$=\frac{18-\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{4}{\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}$

$=\frac{18-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}$

$=\frac{18-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{x+\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}$

$=\frac{18-\sqrt{x}-4\sqrt{x}-8+x+\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}$

$=\frac{18-\sqrt{x}-4\sqrt{x}-8+x+\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}$

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

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

3) P = A.B $=\frac{\sqrt{x}+1}{x-4}.\frac{\sqrt{x}-2}{\sqrt{x}+2}$

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

=$=\frac{\sqrt{x}+1}{{\left(\sqrt{x}+2\right)}^2}=\frac{\sqrt{x}+1}{x+4\sqrt{x}+4}$

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

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

Ta có $\sqrt{x}\ge 0\iff \sqrt{x}+2\ge 2$

$\iff \frac{2}{\sqrt{x}+2}\le \frac{2}{2}\iff -\frac{1}{\sqrt{x}+2}\ge -1$

$\iff 1-\frac{1}{\sqrt{x}+2}\ge 1-1=0$

$$\begin{gathered} \iff {\left(1-\frac{1}{\sqrt{x}+2}\right)}^2\ge 0 \\ \iff -\frac{1}{4}{\left(1-\frac{1}{\sqrt{x}+2}\right)}^2\le 0 \\ \iff -\frac{1}{4}{\left(1-\frac{1}{\sqrt{x}+2}\right)}^2\le 0 \\ \iff P=\frac{1}{4}-\frac{1}{4}{\left(1-\frac{1}{\sqrt{x}+2}\right)}^2\le \frac{1}{4} \end{gathered}$$.

Vậy giá trị lớn nhất của P là $\frac{1}{4}$ . Dấu bằng xảy ra khi $\sqrt{x}=0\iff x=0$.