đk : x >= 0 ; x khác 4 

\(A=\dfrac{2\sqrt{x}-4-\sqrt{x}-2+4}{x-4}=\dfrac{\sqrt{x}-2}{x-4}=\dfrac{1}{\sqrt{x}+2}\)

\(A=\dfrac{2}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}+\dfrac{4}{x-4}\left(đk:x>2\right)\)

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

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

ĐKXĐ: x khác 4; x ≥ 0

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

\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{60}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(4\sqrt{3}+\sqrt{5}\right)^2}\)

\(=2\sqrt{2}-\sqrt{5}-4\sqrt{3}-\sqrt{5}\)

\(=2\sqrt{2}-4\sqrt{3}-2\sqrt{5}\)

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

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

a) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{60}}\)

\(=2\sqrt{2}-\sqrt{5}-4\sqrt{3}+\sqrt{5}\)

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

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

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

=4

a)\(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}=\sqrt{\dfrac{1}{2}\left(16+8\sqrt{3}\right)}-\sqrt{\dfrac{1}{2}\left(16-8\sqrt{3}\right)}\)

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

b)\(=\dfrac{\sqrt{16+2.4\sqrt{5}+5}}{4+\sqrt{5}}.\sqrt{\left(2-\sqrt{5}\right)^2}\)\(=\dfrac{\sqrt{\left(4+\sqrt{5}\right)^2}}{4+\sqrt{5}}\left|2-\sqrt{5}\right|=\sqrt{5}-2\)

a) Ta có: \(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}\)

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

\(=2\sqrt{2}\)

b) Ta có: \(\dfrac{\sqrt{21+8\sqrt{5}}}{4+\sqrt{5}}\cdot\sqrt{9-4\sqrt{5}}\)

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

=16-5=11

mai mk thi rùi cầu cho các bạn trai xinh gái đẹp giúp mk với huhu

`A=1/[\sqrt{3}+1]+1/[\sqrt{3}-1]`

`A=[\sqrt{3}-1+\sqrt{3}+1]/[3-1]`

`A=[2\sqrt{3}]/2=\sqrt{3}`

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

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

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

\(A=\dfrac{2\sqrt{3}}{2}\)

\(A\sqrt{3}\)

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

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

\(A=\dfrac{2\sqrt{3}}{2}\)

\(A=\sqrt{3}\)

`A=((3sqrtx+6)/(x-4)+sqrtx/(sqrtx-2)):(x-9)/(sqrtx-3)(x>=0,x ne 4,x ne 9)`

`=((3(sqrtx+2))/((sqrtx-2)(sqrtx+2))+sqrtx/(sqrtx-2)):((sqrtx-3)(sqrtx+3))/(sqrtx-3)`

`=(3/(sqrtx-2)+sqrtx/(sqrtx-2)):(sqrtx+3)`

`=(sqrtx+3)/(sqrtx-2)*1/(sqrtx+3)`

`=1/(sqrtx-2)`

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

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

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

Ta có: \(A=\left(\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{x-9}{\sqrt{x}-3}\)

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

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

\(a,=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}\)

\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

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

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

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

\(b,=\sqrt{6-2\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}\)

\(=\sqrt{6-2\sqrt{3+\sqrt{12}+1}}\)

\(=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}}\)

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

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

\(c,=\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{4+2.2\sqrt{3}+3}}}\)

\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)

\(=\sqrt{\sqrt{3}+\sqrt{25-2.5\sqrt{3}+3}}\)

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

\(d,=\sqrt{23-6\sqrt{10+4\sqrt{2-2\sqrt{2}+1}}}\)

\(=\sqrt{23-6\sqrt{6+4\sqrt{2}}}\)

\(=\sqrt{23-6\sqrt{4+2.2\sqrt{2}+2}}\)

\(=\sqrt{23-6\sqrt{\left(2+\sqrt{2}\right)^2}}\)

\(=\sqrt{23-12-6\sqrt{2}}=\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)

a) Ta có: \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

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

\(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)

b) Ta có: \(\sqrt{6-2\sqrt{3+\sqrt{13+4\sqrt{3}}}}\)

\(=\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)

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

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

c) Ta có: \(\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)

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

\(=\sqrt{5}\)

d) Ta có: \(\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}\)

\(=\sqrt{23-6\sqrt{10+4\left(\sqrt{2}-1\right)}}\)

\(=\sqrt{23-6\sqrt{6-4\sqrt{2}}}\)

\(=\sqrt{23-6\left(2-\sqrt{2}\right)}\)

\(=\sqrt{11+6\sqrt{2}}\)

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

\(A=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{x-25}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)