\(\frac{18}{\sqrt{7}-1}=\frac{18\left(\sqrt{7}+1\right)}{\left(\sqrt{7}-1\right)\left(\sqrt{7}+1\right)}=\frac{18\left(\sqrt{7}+1\right)}{7-1}=3\sqrt{7}+3\)

Bài này dùng bđt phụ dạng \(\frac{1}{n+n_1+n_2+...+n_m}\le\frac{1}{m^2}\left(\frac{1}{n}+\frac{1}{n_1}+\frac{1}{n_2}+...+\frac{1}{n_m}\right)\)với m = 12 

nhưng bị thiếu mất giả thiết rồi:(

\(q,\frac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

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

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

\(\frac{-2}{\sqrt{2}}\)

\(-\sqrt{2}\)

\(s,\sqrt{5-\sqrt{\left(2\sqrt{3}^2\right)+4\sqrt{3}+1}}+\sqrt{3+\sqrt{\left(2\sqrt{3}^2\right)+4\sqrt{3}+1}}\)

\(\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}\)

\(\sqrt{5-2\sqrt{3}-1}+\sqrt{3+2\sqrt{3}+1}\)

\(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

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

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

\(2\sqrt{3}\)

s , \(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)

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

\(=\sqrt{5-2\sqrt{3}-1}+\sqrt{3+2\sqrt{3}+1}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

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

\(Q=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

\(\sqrt{2}Q=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)

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

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

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

\(=\sqrt{3}-1-\sqrt{3}-1\)

\(=-2\)

\(\Rightarrow Q=-\sqrt{2}\)

\(ĐKXĐ:\hept{\begin{cases}x\ge0\\y\ge1\\z\ge2\end{cases}}\)

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

\(\Leftrightarrow2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)

\(\Leftrightarrow x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)

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

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)

Với \(x\ge0\); \(y\ge1\); \(z\ge2\)ta có:

\(\left(\sqrt{x}-1\right)^2\ge0\); \(\left(\sqrt{y-1}-1\right)^2\ge0\); \(\left(\sqrt{z-2}-1\right)^2\ge0\)

\(\Rightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}-1=0\\\sqrt{y-1}-1=0\\\sqrt{z-2}-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y-1=1\\z-2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( thỏa mãn ĐKXĐ )

Vậy cặp giá trị \(\left(x;y;z\right)\)thỏa mãn đề bài là \(\left(1;2;3\right)\)

a, \(A=\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}+1}=\frac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)ĐK : \(x\ne1;x\ge0\)

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

b, Thay \(x=\frac{9}{4}\Rightarrow\sqrt{x}=\frac{3}{2}\)vào biểu thức A ta được 

\(\frac{\frac{3}{2}}{\frac{3}{2}-1}=\frac{\frac{3}{2}}{\frac{1}{2}}=3\)Vậy với x = 9/4 thì A = 3 

c, Ta có : \(A=\frac{9}{4}\Rightarrow\frac{\sqrt{x}}{\sqrt{x}-1}=\frac{9}{4}\Rightarrow4\sqrt{x}=9\sqrt{x}-9\)

\(\Leftrightarrow5\sqrt{x}=9\Leftrightarrow\sqrt{x}=\frac{9}{5}\Leftrightarrow x=\frac{81}{25}\)

Vậy với A = 9/4 thì x = 81/25 

\(ĐKXĐ=x\ne1;x>0\)

\(A=\frac{\sqrt{x}^3+1}{x-1}-\frac{x-1}{\sqrt{x}+1}\)

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

\(A=\frac{\sqrt{x}^3+1-\sqrt{x}^3+\sqrt{x}+x-1}{x-1}\)

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

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

\(A=\frac{\sqrt{x}}{\sqrt{x}-1}\)

\(b,A=\frac{\sqrt{\frac{9}{4}}}{\sqrt{\frac{9}{4}}-1}=\frac{\frac{3}{2}}{\frac{3}{2}-1}=\frac{3}{\frac{2}{\frac{1}{2}}}=3\)

\(c,\frac{5}{4}=\frac{\sqrt{x}}{\sqrt{x}-1}\)

\(5\sqrt{x}-5=4\sqrt{x}\)

\(\sqrt{x}=5< =>x=25\)

a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x-1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

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

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

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

b) Ta có: \(x=\frac{4}{9}\)thỏa mãn ĐKXĐ

  \(\Rightarrow\)Thay \(x=\frac{4}{9}\)vào biểu thức A ta có:

\(A=\frac{\sqrt{\frac{4}{9}}+2}{\sqrt{\frac{4}{9}}-1}=\frac{\frac{2}{3}+2}{\frac{2}{3}-1}=\frac{\frac{8}{3}}{-\frac{1}{3}}=-8\)

c) Ta có: \(A=\frac{5}{4}\)\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\)

\(\Leftrightarrow4\left(\sqrt{x}+2\right)=5\left(\sqrt{x}-1\right)\)\(\Leftrightarrow4\sqrt{x}+8=5\sqrt{x}-5\)

\(\Leftrightarrow\sqrt{x}=13\)\(\Leftrightarrow x=169\)( thỏa mãn ĐKXĐ )

 Vậy \(x=169\)

\(a,ĐKXĐ:x\ne1,x>0\)

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

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

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

với \(x=\frac{4}{9}\)

\(< =>A=\frac{2+\sqrt{\frac{4}{9}}}{\sqrt{\frac{4}{9}}-1}\)

\(A=\frac{2+\frac{2}{3}}{\frac{2}{3}-1}=\frac{\frac{8}{3}}{\frac{-1}{3}}=-8\)

\(c,\frac{5}{4}=\frac{2+\sqrt{x}}{\sqrt{x}-1}\)

\(5\sqrt{x}-5=8+4\sqrt{x}\)

\(\sqrt{x}=13< =>x=169\)