\(đkcđ\Leftrightarrow x\ge0\)

\(B=\frac{x+5}{\sqrt{x}+2}=\frac{x-4+9}{\sqrt{x}+2}=\frac{x-4}{\sqrt{x}+2}+\frac{9}{\sqrt{x}+2}.\)

\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\frac{9}{\sqrt{x}+2}=\sqrt{x}-2+\frac{9}{\sqrt{x}+2}\)

\(=\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-4\)

Áp dụng bđt Cô - si cho hai số dương \(\sqrt{x}+2\)và \(\frac{9}{\sqrt{x}+2}\), ta có :

\(\sqrt{x}+2+\frac{9}{\sqrt{x}+2}\ge2\sqrt{\frac{\left(\sqrt{x}+2\right).9}{\sqrt{x}+2}}\)

\(\Rightarrow\sqrt{x}+2+\frac{9}{\sqrt{x}+2}\ge2.3\)

\(\Rightarrow\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-4\ge6-4\)

\(\Rightarrow\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-4\ge2\)

Hay \(B_{min}=2\)\(\Leftrightarrow\sqrt{x}+2=\frac{9}{\sqrt{x}+2}\)

\(\Rightarrow\sqrt{x}+2-\frac{9}{\sqrt{x}+2}=0\)

\(\Rightarrow\frac{\left(\sqrt{x}+2\right)^2-9}{\sqrt{x}+2}=0\)

\(\Rightarrow\left(\sqrt{x}+2\right)^2-3^2=0\)

\(\Rightarrow\left(\sqrt{x}+2-3\right)\left(\sqrt{x}+2+3\right)=0\)

\(\Rightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+5\right)=0\)

Vì \(\sqrt{x}+5>0\Rightarrow\sqrt{x}-1=0\)

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

\(KL:B_{min}=2\Leftrightarrow x=1\)

     2\(\sqrt{\dfrac{16}{3}}\)  - 3\(\sqrt{\dfrac{1}{27}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{8}{\sqrt{3}}\) - \(\dfrac{3}{3\sqrt{3}}\)  - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{8}{\sqrt{3}}\) - \(\dfrac{1}{\sqrt{3}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{16}{2\sqrt{3}}\) - \(\dfrac{2}{2\sqrt{3}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{11}{2\sqrt{3}}\)

= \(\dfrac{11\sqrt{3}}{6}\)

f, 2\(\sqrt{\dfrac{1}{2}}\)- \(\dfrac{2}{\sqrt{2}}\) + \(\dfrac{5}{2\sqrt{2}}\)

= \(\dfrac{2}{\sqrt{2}}\) - \(\dfrac{2}{\sqrt{2}}\) + \(\dfrac{5}{2\sqrt{2}}\)

= \(\dfrac{5}{2\sqrt{2}}\)

= \(\dfrac{5\sqrt{2}}{4}\)

(1 + \(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\)).(1- \(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\)) 

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

= \(\dfrac{2}{\sqrt{3}-1}\).\(\dfrac{-2}{\sqrt{3}+1}\)

= \(\dfrac{-4}{3-1}\)

= \(\dfrac{-4}{2}\)

= -2

   \(\dfrac{2}{\sqrt{6}-2}+\dfrac{2}{\sqrt{6}+2}+\dfrac{5}{\sqrt{6}}\)

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

= \(\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{6-4}\) + \(\dfrac{5\sqrt{6}}{6}\)

= \(\dfrac{4\sqrt{6}}{2}\) + \(\dfrac{5\sqrt{6}}{6}\)

= \(\dfrac{12\sqrt{6}+5\sqrt{6}}{6}\)

= \(\dfrac{17\sqrt{6}}{6}\)

ĐKXĐ: x>0

\(\dfrac{x+3}{\sqrt{x}}=\dfrac{x}{\sqrt{x}}+\dfrac{3}{\sqrt{x}}=\sqrt{x}+\dfrac{3}{\sqrt{x}}>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{3}{\sqrt{x}}}=2\sqrt{3}\)

Dấu '=' xảy ra khi \(\left(\sqrt{x}\right)^2=3\)

=>x=3

\(P=\sqrt[]{x}+\dfrac{3}{\sqrt[]{x}-1}\left(x>1\right)\)

\(P=\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}+1\)

Áp dụng bất đẳng thức Cauchy cho 2 số \(\sqrt[]{x}-1;\dfrac{3}{\sqrt[]{x}-1}\) ta được :

\(\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}\ge2\sqrt[]{\sqrt[]{x}-1.\dfrac{3}{\sqrt[]{x}-1}}\)

\(\Rightarrow\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}\ge2\sqrt[]{3}\)

\(\Rightarrow P=\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}+1\ge2\sqrt[]{3}+1\)

\(\Rightarrow Min\left(P\right)=2\sqrt[]{3}+1\)

sorry mn cho e sửa lại đề ạ

tìm gtln của p ạ