\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)

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

\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)

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

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

\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)

\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)

\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)

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

\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

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

\(=\frac{3}{\sqrt{x}-6}\)

Bài 1: Tính

a) Ta có: \(\frac{\sqrt{6+\sqrt{11}}-\sqrt{7-\sqrt{33}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{14-2\sqrt{33}}}{\sqrt{12}+2}\)

\(=\frac{\sqrt{11+2\cdot\sqrt{11}\cdot1+1}-\sqrt{11-2\cdot\sqrt{11}\cdot\sqrt{3}+3}}{2\sqrt{3}+2}\)

\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}}{2\sqrt{3}+2}\)

\(=\frac{\left|\sqrt{11}+1\right|-\left|\sqrt{11}-\sqrt{3}\right|}{2\left(\sqrt{3}+1\right)}\)

\(=\frac{\sqrt{11}+1-\left(\sqrt{11}-\sqrt{3}\right)}{2\left(1+\sqrt{3}\right)}\)(Vì \(\left\{{}\begin{matrix}\sqrt{11}>1>0\\\sqrt{11}>\sqrt{3}\end{matrix}\right.\))

\(=\frac{\sqrt{11}+1-\sqrt{11}+\sqrt{3}}{2\left(1+\sqrt{3}\right)}\)

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

b) Ta có: \(\frac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\frac{2}{4+\sqrt{15}}-\frac{5\sqrt{5}+3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)

\(=\frac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\frac{2}{4+\sqrt{15}}-\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(8-\sqrt{15}\right)}{\sqrt{5}+\sqrt{3}}\)

\(=\sqrt{15}+\frac{2}{4+\sqrt{15}}-\left(8-\sqrt{15}\right)\)

\(=\sqrt{15}+\frac{2}{4+\sqrt{15}}-8+\sqrt{15}\)

\(=2\sqrt{15}-8+\frac{2}{4+\sqrt{15}}\)

\(=\frac{2\sqrt{15}\left(4+\sqrt{15}\right)}{4+\sqrt{15}}-\frac{8\left(4+\sqrt{15}\right)}{4+\sqrt{15}}+\frac{2}{4+\sqrt{15}}\)

\(=\frac{8\sqrt{15}+30-32-8\sqrt{15}+2}{4+\sqrt{15}}\)

\(=\frac{0}{4+\sqrt{15}}=0\)

Bài 2: Rút gọn

Ta có: \(B=\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\left(\frac{1+\sqrt{a}}{a-1}\right)^2\)

\(=\left(\frac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)}{1+\sqrt{a}}-\sqrt{a}\right)\cdot\left(\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)^2\)

\(=\left(1-\sqrt{a}+a-\sqrt{a}\right)\cdot\left(\frac{1}{\sqrt{a}-1}\right)^2\)

\(=\left(a-2\sqrt{a}+1\right)\cdot\frac{1}{\left(\sqrt{a}-1\right)^2}\)

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

Bài 3:

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

b) Ta có: \(A=\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{3-3\sqrt{x}}{x-5\sqrt{x}+6}\)

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

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

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

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

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

c) Để A<-1 thì A+1<0

\(\Leftrightarrow\frac{1}{3-\sqrt{x}}+1< 0\)

\(\Leftrightarrow\frac{-1}{\sqrt{x}-3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\frac{-1+\sqrt{x}-3}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-4>0\\\sqrt{x}-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-4< 0\\\sqrt{x}-3>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>4\\\sqrt{x}< 3\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 4\\\sqrt{x}>3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\Leftrightarrow9< x< 16\)

\(A=\left(1-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\right):\left(\frac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)-\left(9-a\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\right)\)

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

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

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

Để \(A+\left|A\right|\ne0\Rightarrow\left|A\right|\ne-A\Rightarrow A>0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{a}< 2\\\sqrt{a}>3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a< 4\\a>9\end{matrix}\right.\)

Kết hợp điều kiện \(\Rightarrow\left[{}\begin{matrix}0\le a< 4\\a>9\end{matrix}\right.\)

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

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

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

b: Để A<0 thì \(\sqrt{x}-2< 0\)

hay 0<x<4

a) \(A=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)\)

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

\(A=\frac{\sqrt{3}+1}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\frac{5+3\sqrt{5}}{\sqrt{5}}\)

\(A=1\)

b) Ta có:

\(B=\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\) ( x >= 0, x khác 9 )

\(B=\frac{3+\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}+\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)

\(B=\frac{3+\sqrt{x}+3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)

\(B=\frac{3+\sqrt{x}+3\sqrt{x}+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)

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

\(B=\frac{4\left(3+\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)

\(B=\frac{4}{3-\sqrt{x}}\)

Để B > A

\(\Rightarrow\frac{4}{3-\sqrt{x}}>1\)

\(\Rightarrow4>3-\sqrt{x}\)

\(\Rightarrow4-3+\sqrt{x}>0\)

\(\Rightarrow1+\sqrt{x}>0\)

\(\Rightarrow\sqrt{x}>-1\)

\(\Rightarrow x>1\)

a) A=\(\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)\)

\(=\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}+\frac{\sqrt{5}\cdot\left(\sqrt{5}+3\right)}{\sqrt{5}}\)

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

\(=\frac{\sqrt{3}+1}{\sqrt{3}+1}+0=1\)

b) B=\(\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\)

\(=\frac{3+\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}+\frac{x+9}{9-x}\)

\(=\frac{3+\sqrt{x}+3\sqrt{x}-x}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}\)

\(=\frac{4\text{​​}\sqrt{x}+12}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}\)

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

\(=\frac{4}{3-\sqrt{x}}\)

\(B>A \Leftrightarrow\frac{4}{3-\sqrt{x}}>1\)

các giá trị của x là \(\left\{x\in R\backslash0\le x\le9\right\}\)

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

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

\(=\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

b. Để \(A< 1\Rightarrow\frac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\Rightarrow\frac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\Rightarrow\frac{4}{\sqrt{x}-3}< 0\)

\(\Rightarrow\sqrt{x}-3< 0\Rightarrow0\le x< 9\)

Kết hợp đk thì \(0\le x< 9\)và \(x\ne4\)thì \(A< 1\)

\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{x-2}-\frac{2\sqrt{x}+1}{3\sqrt{x}}\)