\(a,\)\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne9;x\ne25\end{cases}}\)

\(P=\frac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}\)\(-\frac{\sqrt{x}+15}{\sqrt{x}-3}-\frac{3\sqrt{x}-1}{5-\sqrt{x}}\)

\(=\frac{8\sqrt{x}-x-31}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)\(-\frac{\sqrt{x}+15}{\sqrt{x}-3}+\frac{3\sqrt{x}-1}{\sqrt{x}-5}\)

\(=\frac{8\sqrt{x}-x-31}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}-\)\(\frac{\left(\sqrt{x}+15\right)\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)\(+\frac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)

\(=\frac{8\sqrt{x}-x-31-x-10\sqrt{x}+75+3x-10\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)

\(=\frac{x-12\sqrt{x}+47}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)

\(\Rightarrow\)Sai đề không cậu ưi 

a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

\(M=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{x-9}\)

\(=\frac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}.\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)

b) Ta có: \(x=\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}=\sqrt{\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}\)

\(=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{\sqrt{3}-\left|\sqrt{3}-1\right|}\)

\(=\sqrt{\sqrt{3}-\sqrt{3}+1}=\sqrt{1}=1\)( thỏa mãn ĐKXĐ )

Thay \(x=1\)vào M ta được:

\(M=\frac{3\sqrt{1}}{\sqrt{1}-3}=\frac{3}{1-3}=\frac{-3}{2}\)

c) \(M=\frac{3\sqrt{x}}{\sqrt{x}-3}=\frac{3\sqrt{x}-9+9}{\sqrt{x}-3}=\frac{3\left(\sqrt{x}-3\right)+9}{\sqrt{x}-3}=3+\frac{9}{\sqrt{x}-3}\)

Vì \(x\inℕ\)\(\Rightarrow\)Để M là số tự nhiên thì \(\frac{9}{\sqrt{x}-3}\inℕ\)

\(\Rightarrow9⋮\left(\sqrt{x}-3\right)\)\(\Rightarrow\sqrt{x}-3\inƯ\left(9\right)\)(1)

Vì \(x\ge0\)\(\Rightarrow\sqrt{x}\ge0\)\(\Rightarrow\sqrt{x}-3\ge-3\)(2)

Từ (1) và (2) \(\Rightarrow\sqrt{x}-3\in\left\{-3;-1;1;3;9\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{0;2;4;6;12\right\}\)\(\Rightarrow x\in\left\{0;4;16;36;144\right\}\)( thỏa mãn ĐKXĐ )

Thử lại với \(x=4\)ta thấy M không là số tự nhiên

Vậy \(x\in\left\{0;16;36;144\right\}\)

đkxđ là \(x\ne1;x>0\)

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

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

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

gtnn \(x-\sqrt{x}+1=x-\frac{1}{2}.2.\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

gtnn 3/4

ý c bạn tự làm nha mk chịu

mình cảm ơn bạn nha 

\(A=\left(\frac{1+\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\frac{1-\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\right).\sqrt{3}\)

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

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

Để \(A=\frac{B}{6}\Leftrightarrow B=6A\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}}=-18\)

\(\Rightarrow\sqrt{x}-1=-18\sqrt{x}\Rightarrow\sqrt{x}=\frac{1}{19}\Rightarrow x=\frac{1}{361}\)

a.\(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\sqrt{x}-1\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b.\(Q=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)

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

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

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

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

c.Để \(Q=\frac{2}{7}\) thì \(\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{2}{7}\)

\(\Leftrightarrow7\sqrt{x}=2x+2\sqrt{x}+2\)

\(\Leftrightarrow2x-5\sqrt{x}+2=0\)

\(\Leftrightarrow2x-4\sqrt{x}-\sqrt{x}+2=0\)

\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\2\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{1}{4}\end{matrix}\right.\)(Thỏa mãn đkxđ)