a: \(=\dfrac{x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

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

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

b: khi x=6-2căn 5 thì \(P=\dfrac{6-2\sqrt{5}-3\sqrt{5}+3-5}{\left(\sqrt{5}-3\right)\left(\sqrt{5}-4\right)\cdot\sqrt{5}}\)

\(=\dfrac{-5\sqrt{5}+4}{\sqrt{5}\left(\sqrt{5}-3\right)\left(\sqrt{5}-4\right)}\)

a) \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+3}\right)\) (ĐK: \(x>0;x\ne1\)) 

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

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

\(A=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}:\dfrac{2\sqrt{x}}{\sqrt{x}+3}\)

\(A=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+3}{2\sqrt{x}}\)

\(A=\dfrac{\sqrt{x}+3}{\sqrt{x}}\)

b) Ta có: \(x=\dfrac{1}{6-2\sqrt{5}}=\dfrac{1}{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot1+1^2}=\dfrac{1}{\left(\sqrt{5}-1\right)^2}=\left(\dfrac{1}{\sqrt{5}-1}\right)^2\)

Thay vào A ta có:

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

c) Ta có: \(\dfrac{\sqrt{x}+3}{\sqrt{x}}=1+\dfrac{3}{\sqrt{x}}\)

\(\Rightarrow\sqrt{x}\in\left\{1;3\right\}\)

\(\Rightarrow x\in\left\{1;9\right\}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)

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

\(=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+3}{6}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+3}{6}=\dfrac{\sqrt{x}+3}{3}\)

b: Khi \(x=\dfrac{1}{6-2\sqrt{5}}=\dfrac{6+2\sqrt{5}}{16}=\left(\dfrac{\sqrt{5}+1}{4}\right)^2\) thì \(A=\dfrac{\dfrac{\sqrt{5}+1}{4}+3}{3}=\dfrac{\sqrt{5}+1+12}{12}=\dfrac{13+\sqrt{5}}{12}\)

c: A là số nguyên

=>\(\sqrt{x}+3⋮3\)

=>\(\sqrt{x}⋮3\)

=>\(x=k^2\);\(k\in Z\)

Kết hợp ĐKXĐ, ta được: x là số chính phương và x>0 và \(x\ne1\)

1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)

Ta có: \(A=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(1,A=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\left(x\ge0;x\ne4;x\ne9\right)\\ 2,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow0\le x< 9\)

a: Ta có: \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

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

\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3}{\sqrt{x}+3}\)

a: \(A=\dfrac{2x-6\sqrt{x}+\sqrt{x}-3-2x+4\sqrt{x}+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{3x-3\sqrt{x}-\sqrt{x}-4}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1}{3x-4\sqrt{x}-4}\)

\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{3x-6\sqrt{x}+2\sqrt{x}-4}{\sqrt{x}-1}=\dfrac{3\sqrt{x}+2}{\sqrt{x}-1}\)

b: Để A<2 thì \(\dfrac{3\sqrt{x}+2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)}< 0\)

=>x<1

=>x<1

1) ĐKXĐ: \(x\notin\left\{0;1\right\}\)

2) Ta có: \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)

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

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

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

ĐKXĐ: \(x\ge0;x\ne1\)

\(P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

b.

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

Do \(\left\{{}\begin{matrix}2\sqrt{x}\ge0\\\sqrt{x}+1>0\end{matrix}\right.\) \(\Rightarrow\dfrac{2\sqrt{x}}{\sqrt{x}+1}\ge0\)

\(\Rightarrow P\ge-1\)

\(P_{min}=-1\) khi \(x=0\)

a) Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

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

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)

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

Câu 3: 

Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)

\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)

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

\(=a-2\sqrt{a}\)

1: \(B=\dfrac{2\sqrt{x}-x-2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)

\(=\dfrac{-x}{\left(\sqrt{x}-2\right)\cdot\sqrt{x}}\)

\(=\dfrac{-\sqrt{x}}{\sqrt{x}-2}\)