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

\(đkxđ\Leftrightarrow\hept{\begin{cases}x^2-x+1\ge0\\x^2-x+1\ne0\end{cases}\Rightarrow x^2-x+1>0}\)

Mà \(x^2-x+1=x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)với \(\forall x\)

\(\Rightarrow\)Biểu thức luôn được xác định với mọi x 

\(b,\sqrt{\frac{2x-1}{x+3}}\)

\(Đk:\)\(x+3\ne0\Rightarrow x\ne-3\)

Và \(\frac{2x-1}{x+3}\ge0\)

Khi \(\frac{2x-1}{x+3}=0\Rightarrow2x-1=0\)

\(\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)

Khi \(\frac{2x-1}{x+3}>0\)\(\Rightarrow\orbr{\begin{cases}2x-1>0;x+3>0\\2x-1< 0;x+3< 0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>\frac{1}{2};x>-3\\x< \frac{1}{2};x< -3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x< -3\end{cases}}\)

Vậy căn thức xác định khi \(x\ge\frac{1}{2};x< -3\)

Bài 1 : Với : \(x>0;x\ne1\)

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

Thay vào ta được : \(P=x=25\)

Bài 2 : 

a, Với \(x\ge0;x\ne1\)

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

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

Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)

Bài 3 : \(x\ge0;x\ne1\)

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

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

b, Ta có : \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\Rightarrow4\sqrt{x}+8=5\sqrt{x}-5\)

\(\Leftrightarrow\sqrt{x}=13\Leftrightarrow x=169\)(tmđk )

1) \(\frac{1}{\sqrt{2x-1}}\)có nghĩa khi \(\hept{\begin{cases}2x-1\ge0\\\sqrt{2x-1}\ne0\end{cases}}\)

\(\Leftrightarrow2x-1>0\)

\(\Leftrightarrow x>\frac{1}{2}\)

\(\sqrt{5-x}\)có nghĩa khi \(5-x\ge0\Leftrightarrow x\ge5\)

Vậy \(ĐKXĐ:\frac{1}{2}>x\ge5\)

2) \(\sqrt{x-\frac{1}{x}}\)có nghĩa khi \(\hept{\begin{cases}x-\frac{1}{x}\ge0\\x>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{x^2}{x}-\frac{1}{x}\ge0\\x>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{x^2-1}{x}\ge0\\x>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2-1\ge0\\x>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2\ge1\\x>0\end{cases}}\)

Vậy \(ĐKXĐ:x\ge1\)

3) \(\sqrt{2x-1}\)có nghĩa khi \(2x-1\ge0\) \(\Leftrightarrow x\ge\frac{1}{2}\)

\(\sqrt{4-x^2}\)có nghĩa khi \(4-x^2\ge0\Leftrightarrow x^2\le4\Leftrightarrow x\le2\)

Vậy \(ĐKXĐ:\frac{1}{2}\le x\le2\)

4) \(\sqrt{x^2-1}\)có nghĩa khi \(x^2-1\ge0\Leftrightarrow x^2\ge1\Leftrightarrow x\ge1\)

\(\sqrt{9-x^2}\)có nghĩa khi \(9-x^2\ge0\Leftrightarrow x^2\le9\Leftrightarrow x\le3\)

Vậy \(ĐKXĐ:1\le x\le3\)

\(a,\)\(\frac{1}{1-\sqrt{x^2-3}}\)

\(đkxđ\Leftrightarrow\orbr{\begin{cases}x^2-3\ge0\\x^2-3\ne1\end{cases}}\).

\(x^2-3\ne1\)\(\Rightarrow x^2\ne4\)\(\Rightarrow x\ne\pm2\)

\(x^2-3\ge0\)\(\Rightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\ge0\)

Chia trường hợp ra làm nốt nhé 

....

\(b,\)\(\frac{x-1}{2-\sqrt{3x+1}}\)

\(đkxđ\Leftrightarrow\orbr{\begin{cases}3x+1\ge0\\\sqrt{3x+1}\ne2\end{cases}}\)

\(3x+1\ge0\)\(\Rightarrow3x\ge-1\)

\(\Rightarrow x\ge\frac{-1}{3}\)

\(\sqrt{3x+1}\ne2\)\(\Rightarrow|3x+1|\ne4\)\(\Rightarrow\hept{\begin{cases}3x-1\ne4\\3x-1\ne-4\end{cases}\Rightarrow\hept{\begin{cases}3x\ne5\\3x\ne-3\end{cases}\Rightarrow}\hept{\begin{cases}x\ne\frac{5}{3}\\x\ne-1\end{cases}}}\)

\(\Rightarrow x\ge-\frac{1}{3}\)và \(x\ne\frac{5}{3}\)

\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x\ne4\right)\)

\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

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

\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

b) Với  \(x=3\)( thỏa mãn ĐKXĐ ) ta có  \(P=\frac{3\sqrt{3}}{\sqrt{3}+2}=-9+6\sqrt{3}\)

c) A ở đâu ???? '-'