a: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)

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

\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

b: \(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

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

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

\(=3+\sqrt{5}+3-\sqrt{5}=6\)

c: \(\dfrac{3}{2\sqrt{3}+3}+\dfrac{3}{2\sqrt{3}-3}\)

\(=\dfrac{3\left(2\sqrt{3}-3\right)+3\left(2\sqrt{3}+3\right)}{12-9}\)

\(=2\sqrt{3}-3+2\sqrt{3}+3=4\sqrt{3}\)

d: \(\sqrt{\left(\sqrt{3}+4\right)\cdot\sqrt{19-8\sqrt{3}}+3}\)

\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\sqrt{\left(4-\sqrt{3}\right)^2}+3}\)

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

\(=\sqrt{16-3+3}=\sqrt{16}=4\)

e: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)

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

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

\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

\(B=\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(B=\left|\sqrt{5}+2\right|+\left|\sqrt{5}-2\right|\)

\(B=\sqrt{5}+2+\sqrt{5}-2\)

\(B=2\sqrt{5}\)

\(A=\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}\)

\(A=\left(\dfrac{\sqrt{12}-\sqrt{6}}{2\sqrt{2}-2}-\dfrac{6\sqrt{6}}{3}\right).\dfrac{1}{\sqrt{6}}\)

\(A=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)

\(A=\left(\sqrt{6}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)

\(A=-\sqrt{6}.\dfrac{1}{\sqrt{6}}\)

\(A=-1\)

a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}=14\sqrt{2}-9\sqrt{2}+2\sqrt{2}=7\sqrt{2}\)

b) \(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}=5\sqrt{10}+10-5\sqrt{10}=10\)

c) \(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}=6-5\sqrt{6}-6=5\sqrt{6}\)

d) \(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}=12\sqrt{3}+6\sqrt{3}-3\sqrt{3}=15\sqrt{3}\)

e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}=2\sqrt{3}+3\sqrt{3}=\left(\sqrt{3}+1\right)=4\sqrt{3}-1\)

f) \(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}.\dfrac{2}{\sqrt{2}+1}=2\sqrt{2}-\left(12-6\sqrt{2}\right)=8\sqrt{2}-12\)

1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)

\(\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}+\sqrt{13-4\sqrt{3}}-\sqrt{22+12\sqrt{2}}\)

\(=\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}+\sqrt{12-2\cdot2\sqrt{3}+1}-\sqrt{18+4\cdot3\sqrt{2}+4}\)

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

\(=3\sqrt{2}-2\sqrt{3}+2\sqrt{3}-1-3\sqrt{2}-2\)

\(=-3\)

a) ĐKXĐ có thêm \(x\ne4\)

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

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

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

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

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

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

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

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

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

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

2:

a: =căn 17-4-căn 17=-4

b: =5-2căn 3-2căn 3=5-4căn 3

1:

a: =>|x+1|=-x

=>x<=0 và (x+1)^2=x^2

=>x<=0 và (x+1+x)(x+1-x)=0

=>x=-1/2

a) Ta có: \(\sqrt{27\cdot48\left(1-a^2\right)}\)

\(=\sqrt{3^4\cdot4^2\cdot\left(1-a^2\right)}\)

\(=36\sqrt{1-a^2}\)

c) Ta có: \(\sqrt{5a}\cdot\sqrt{45a}-3a\)

\(=15a-3a=12a\)

b) Ta có: \(B=\dfrac{1}{a-b}\cdot\sqrt{a^4\cdot\left(a-b\right)^2}\)

\(=\dfrac{1}{a-b}\cdot a^2\cdot\left(a-b\right)\)

\(=a^2\)

d) Ta có: \(D=\left(3-a\right)^2-\sqrt{0.2}\cdot\sqrt{180a^2}\)

\(=a^2-6a+9-\sqrt{36a^2}\)

\(=a^2-6a+9-\left|6a\right|\)

\(=\left[{}\begin{matrix}a^2-6a+9-6a\left(a\ge0\right)\\a^2-6a+9+6a\left(a< 0\right)\end{matrix}\right.\)

\(=\left[{}\begin{matrix}a^2-12a+9\\a^2+9\end{matrix}\right.\)

\(A=9.4\left|1-a\right|=36\left(a-1\right)\) (a>1)

\(B=\dfrac{a^2\left|a-b\right|}{a-b}=\dfrac{a^2\left(a-b\right)}{a-b}=a^2\) (a>b)

\(C=5.3\left|a\right|-3a=15a-3a=12a\)

\(D=9-6a+a^2-6\left|a\right|=\left[{}\begin{matrix}a^2-12a+9\left(a\ge0\right)\\a^2+9\left(a< 0\right)\end{matrix}\right.\) 

`A=\sqrt{6-2\sqrt{5}}`

`A=\sqrt{(\sqrt{5}-1)^2}`

`A=\sqrt{5}-1`

_________

`B=\sqrt{4-\sqrt{12}}=\sqrt{4-2\sqrt{3}}`

`B=\sqrt{(\sqrt{3}-1)^2}`

`B=\sqrt{3}-1`

_________

`C=\sqrt{19-8\sqrt{3}}`

`C=\sqrt{(4-\sqrt{3})^2}`

`C=4-\sqrt{3}`

_________

`D=\sqrt{5-2\sqrt{6}}`

`D=\sqrt{(\sqrt{3}-\sqrt{2})^2}`

`D=\sqrt{3}-\sqrt{2}`

\(A=\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5}^2-2\sqrt{5}+1^2}=\sqrt{ \left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)

\(B=\sqrt{4-\sqrt{12}}=\sqrt{4-\sqrt{4.3}}=\sqrt{4-2\sqrt{3}}=\sqrt{\sqrt{3^2}-2\sqrt{3}+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

\(C=\sqrt{19-8\sqrt{3}}=\sqrt{19-2.4.\sqrt{3}}\sqrt{\sqrt{3}^2-2.4.\sqrt{3}+4^2}=\sqrt{\left(\sqrt{3}-4\right)^2}=\sqrt{3}-4\)

\(D=\sqrt{5-2\sqrt{6}}=\sqrt{5-2.\sqrt{2}.\sqrt{3}}=\sqrt{\sqrt{3}^2-2.\sqrt{2}.\sqrt{3}+\sqrt{2^2}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)