ta có :

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

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

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

Mình gửi câu hỏi

ok giúp mình có 1 tiick nha

Dễ thấy P>0 nên ta có :

\(\frac{1}{P}=\frac{a+\sqrt{a}+1}{5\sqrt{a}+1}=\frac{1}{5}\left(\sqrt{a}+\frac{4}{5}\right)+\frac{21}{25\left(5\sqrt{a}+1\right)}=\left(\frac{5\sqrt{a}+1}{25}\right)+\frac{21}{25\left(5\sqrt{a}+1\right)}+\frac{3}{25}\)

\(\ge\frac{2\sqrt{21}}{25}+\frac{3}{25}=\frac{3+2\sqrt{21}}{25}\)

\(\Rightarrow P\le\frac{25}{3+2\sqrt{21}}\)là GTLN của P

còn P không có giá trị nhỏ nhất nhé

b, bài này theo mình nghĩ chỉ có GTLN thôi, nếu có GTNN thì bày mình nhé :))  \(P=-\frac{3\sqrt{x}}{\sqrt{x}+3}=\frac{-3\left(\sqrt{x}+3\right)+9}{\sqrt{x}+3}=-3+\frac{9}{\sqrt{x}+3}\) 

Ta có : \(\sqrt{x}+3\ge3\Rightarrow\frac{9}{\sqrt{x}+3}\le\frac{9}{3}=3\)

\(\Rightarrow P=-3+\frac{9}{\sqrt{x}+3}\le-3+3=0\)

Dấu ''='' xảy ra khi x = 0 

Vậy GTLN của P bằng 0 tại x = 0 

e, \(P>-1\Leftrightarrow P+1>0\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}+1>0\)

\(\Leftrightarrow\frac{-3\sqrt{x}+\sqrt{x}+3}{\sqrt{x}+3}>0\Leftrightarrow\frac{3-2\sqrt{x}}{\sqrt{x}+3}>0\Leftrightarrow\frac{2\sqrt{x}-3}{\sqrt{x}+3}< 0\)

\(\Rightarrow2\sqrt{x}-3< 0\Leftrightarrow x< \frac{9}{4}\)

Kết hợp với đk : \(0< x< \frac{9}{4}\)

g, Xét  \(P=-\frac{3\sqrt{x}}{\sqrt{x}+3}\) có \(\sqrt{x}\ge0\Rightarrow-3\sqrt{x}\le0\)mà \(\sqrt{x}+3>0\)

\(\Rightarrow P=-\frac{3\sqrt{x}}{\sqrt{x}+3}\le0\)

Xét \(1=\frac{\sqrt{x}+3}{\sqrt{x}+3}\)mà \(\sqrt{x}+3>0\)

Vậy P < 1 

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

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

\(b.\frac{2x+1}{x-7\sqrt{x}+12}-\frac{\sqrt{x}+3}{\sqrt{x}-4}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}=\frac{2x+1-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}\)

\(=\frac{3x-7\sqrt{x}-6}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}+2}{\sqrt{x}-4}\)

\(c.\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}+\frac{1}{2-\sqrt{x}}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)\(=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-2\right)}\)

giúp mình :(

\(ĐK:x\ge\frac{3}{2}\)

\(3x-8\sqrt{x+14}=2\sqrt{2x-3}-28\)

\(\Leftrightarrow2\sqrt{2x-3}-28-3x+8\sqrt{x+14}=0\)

\(\Leftrightarrow2\cdot\frac{\left(\sqrt{2x-3}-1\right)\left(\sqrt{2x-3}+1\right)}{\sqrt{2x-3}+1}+8\cdot\frac{\left(\sqrt{x+14}-4\right)\left(\sqrt{x+14}+4\right)}{\sqrt{x+14}+4}-3x+6=0\)

\(\Leftrightarrow2\cdot\frac{2x-3-1}{\sqrt{2x-3}+1}+8\cdot\frac{x+14-16}{\sqrt{x+14}+4}-3\left(x-2\right)=0\)

\(\Leftrightarrow\frac{4\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{8\left(x-2\right)}{\sqrt{x+14}+4}-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{4}{\sqrt{2x-3}+1}+\frac{8}{\sqrt{x+14}+4}-3\right)=0\)

th1 : \(x-2=0\Leftrightarrow x=2\left(tm\right)\)

th2 : \(\frac{4}{\sqrt{2x-3}+1}+\frac{8}{\sqrt{x+14}+4}-3=0\)

này thì cũng ra nghiệm = 2 nhưng chưa biết làm ;-;

\(ĐKXĐ:x\ge\frac{3}{2}\)

\(3x-\left(8\sqrt{x+14}-32\right)=\left(2\sqrt{2x-3}-2\right)+6\)

\(3x-\frac{64x+896-1024}{8\sqrt{x+14}+32}=\frac{8x-12-4}{2\sqrt{2x-3}+2}+6\)

\(3x-6-\frac{64 \left(x-2\right)}{8\sqrt{x+14}+32}-\frac{8\left(x-2\right)}{2\sqrt{2x-3}+2}=0\)

\(\left(x-2\right)\left(3-\frac{64}{8\sqrt{x+14}+32}-\frac{8}{2\sqrt{2x-3}+2}\right)=0\)

\(\orbr{\begin{cases}x-2=0\Rightarrow x=2\left(TM\right)\\3-\frac{64}{8\sqrt{x+14}+32}-\frac{8}{2\sqrt{2x-3}+2}=0\end{cases}}\)

CM nốt cái dưới khác 0 nha

\(\)