bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\) 

Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)

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

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

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

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

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

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

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 

A) ĐKXĐ : \(x\ge0\) và \(x\ne4\)

Rút gọn :\(A=\frac{2}{2+\sqrt{x}}+\frac{1}{2-\sqrt{x}}+\frac{4\sqrt{x}}{4-x}\)

            \(A=\frac{2\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{2+\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{4\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)

            \(A=\frac{4-2\sqrt{x}+2+\sqrt{x}+4\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)

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

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

           \(A=\frac{3}{2-\sqrt{x}}\)

b) thay \(x=7+4\sqrt{3}\) vào A 

ta được :\(A=\frac{3}{2-\sqrt{7+4\sqrt{3}}}=\frac{3}{2-2+\sqrt{3}}=\frac{3}{\sqrt{3}}\)

vậy vói \(x=7+4\sqrt{3}\) thì \(A=\frac{3}{\sqrt{3}}\)

c)với\(x\ge0\) và \(x\ne4\)

Để \(A=-\frac{3}{7}\Leftrightarrow\frac{3}{2-\sqrt{x}}=-\frac{3}{7}\)

                        \(\Leftrightarrow3.7=-3\left(2-\sqrt{x}\right)\)

                         \(\Leftrightarrow21=-6+3\sqrt{x}\)

                          \(\Leftrightarrow21+6=3\sqrt{x}\)

                           \(\Leftrightarrow27=3\sqrt{x}\)

                            \(\Leftrightarrow\sqrt{x}=9\)

                           \(\Leftrightarrow x=81\)

Vậy để\(A=-\frac{3}{7}\Leftrightarrow x=81\)

a. ĐKXĐ: 

\(\hept{\begin{cases}\sqrt{x}-1\ne0\\x-\sqrt{x}\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x>0\\x\ne1\end{cases}}}\)

b. ta có \(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)

c. khi \(x=\frac{1}{4}\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow A=\frac{\frac{1}{2}+1}{\frac{1}{2}}=3\)

khi \(x=3+2\sqrt{2}\Rightarrow\sqrt{x}=\sqrt{2}+1\Rightarrow A=\frac{\sqrt{2}+1+1}{\sqrt{2}+1}=\sqrt{2}\)

\(a,ĐKXĐ:A=x\ge0;x\ne1\)

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

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

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

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

\(A=\frac{\sqrt{x}+1}{\sqrt{x}}< =>ĐPCM\)

c,thay \(x=\frac{1}{4}\)vào A

\(c,A=\frac{\sqrt{\frac{1}{4}}+1}{\sqrt{\frac{1}{4}}}\)

\(A=\frac{\frac{1}{2}+1}{\frac{1}{2}}\)

\(A=3\)

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

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

\(x=\left(\sqrt{2}+1\right)^2\)thay x vào A

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

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

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

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

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

\(đk:\)\(\frac{1}{\left(x-3\right)^3}\ne0\)\(\Rightarrow\left(x-3\right)^3\ne0\)\(\Leftrightarrow x\ne3\)

Và \(\frac{1}{\left(x-3\right)}>0\Rightarrow x-3>0\)\(\Rightarrow x>3\)

Vậy để căn thức xác định thì x > 3

\(\sqrt{8x-x^2-15}\)

\(=\sqrt{-\left(x^2-8x+15\right)}\)

\(=\sqrt{-\left(x^2-8x+16-1\right)}\)

\(=\sqrt{-\left[\left(x^2-8x+16\right)-1\right]}\)

\(=\sqrt{-\left(x-4\right)^2+1}\)

\(đk:\)\(-\left(x-4\right)^2+1\ge0\)

\(\Rightarrow\left(x-4\right)^2\le1\)

\(\Rightarrow\orbr{\begin{cases}\left(x-4\right)^2=1\\\left(x-4\right)^2=0\end{cases}}\)

\(\left(x-4\right)^2=1\Rightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}\)

\(\left(x-4\right)^2=0\Rightarrow x=4\)

Vậy căn thức xác định \(\Leftrightarrow x=\left\{3;4;5\right\}\)

mình giúp bài 3 cho 

\(\sqrt{25x-125}-3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=6\left(ĐKXĐ:x\ge5\right)\)

\(< =>\sqrt{25\left(x-5\right)}-3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=6\)

\(< =>\sqrt{25}.\sqrt{x-5}-3\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=6\)

\(< =>5.\sqrt{x-5}-3.\frac{\sqrt{x-5}}{3}-\frac{1}{3}.3.\sqrt{x-5}=6\)

\(< =>5.\sqrt{x-5}-\sqrt{x-5}-\sqrt{x-5}=6\)

\(< =>3\sqrt{x-5}=6< =>\sqrt{x-5}=2\)

\(< =>x-5=4< =>x=4+5=9\left(tmđk\right)\)