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

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

\(Đkxđ:\)

\(\sqrt{x}\ge0\Rightarrow x\ge0\)

\(\sqrt{x}-1\ne0\Rightarrow\sqrt{x}\ne1\Rightarrow x\ne1\)

\(\sqrt{x}\ne0\Rightarrow x\ne0\)

\(\RightarrowĐkxđ:x>0;x\ne1\)

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

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

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

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

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

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

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

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

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

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

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

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

1. \(Q=-\frac{1}{\sqrt{x}-3}\)

để Q nguyên thì \(\sqrt{x}-3\inƯ\left(1\right)=\left(-1;1\right)\)

\(\sqrt{x}-3=-1\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

\(\sqrt{x}-3=1\Rightarrow\sqrt{x}=4\Rightarrow x=16\)

2. \(Q=\frac{\sqrt{x}-3}{\sqrt{x}-1}=1-\frac{2}{\sqrt{x}-1}\)

Để Q nguyên thì \(\sqrt{x}-1\inƯ\left(2\right)=\left(-2;-1;1;2\right)\)

\(\sqrt{x}-1=-2\Rightarrow\sqrt{x}=-1VN\)

\(\sqrt{x}-1=-1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)

\(\sqrt{x}-1=1\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

\(\sqrt{x}-1=2\Rightarrow\sqrt{x}=3\Rightarrow x=9\)

\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{\sqrt{a}-\sqrt{b}}\)

\(=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}-\frac{2b}{\sqrt{a}-\sqrt{b}}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+b\right)}-\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{4b\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2-4b\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}\right)-4\sqrt{a}b-4b\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{2\sqrt{a}.2\sqrt{b}-4\sqrt{a}b-4b\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{4\sqrt{a}\sqrt{b}-4\sqrt{a}b-4b\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{4\sqrt{a}\sqrt{b}\left(1-\sqrt{b}-b\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{2\sqrt{a}\sqrt{b}\left(1-\sqrt{b}-b\right)}{a-b}\)

Đề sai???Phân số thứ 3 nghi là a-b chứ ko phải căn a - căn b????????

\(ĐKXĐ:\hept{\begin{cases}x-4\ne0\\3-\sqrt{x}\ne0\\x\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ne4\\\sqrt{x}\ne3\\x\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ne4\\x\ne9\\x\ge0\end{cases}}\)

Rút gọn

\(D=\left(\frac{x-2\sqrt{x}}{x-4}-1\right):\left(\frac{4-x}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)

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

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

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

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

\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)

\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x-\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)

\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x-\left(x+4\sqrt{x}+4\right)-\left(x-6\sqrt{x}+9\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)

\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x-x^2-4\sqrt{x}-4-x^2+6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)

\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{-2x^2-x-2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)

\(D=\frac{\left(-2\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(-2x^2-x-2\sqrt{x}-9\right)}\)

\(D=\frac{\left(-2\right)\left(\sqrt{x}-3\right)\left(x^2-4\right)}{\left(x-4\right)\left(-2x^2-x-2\sqrt{x}-9\right)}\)

Sai thui nhé !!!!

\(\text{Đat: A=biêu thuc cần tính}\Rightarrow\sqrt{2}A=\sqrt{28+10\sqrt{3}}+\sqrt{4-2\sqrt{3}}\) 

\(\Rightarrow2\sqrt{A}=\sqrt{5^2+2.5\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1^2}\) 

\(\Rightarrow2\sqrt{A}=\sqrt{\left(5+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=4+2\sqrt{3}\Rightarrow A=\sqrt{8}+\sqrt{6}\)

\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

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

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

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

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

\(=2\sqrt{2}\)