\(a,A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)

\(=\sqrt{\left(\sqrt{5}^2+2\sqrt{5}+2\sqrt{2}\cdot\sqrt{5}\right)+\sqrt{2}^2+2\sqrt{2}\cdot1+1^2}\)

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

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

\(=\sqrt{5}+\sqrt{2}+1\)

\(b,B=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\frac{3\cdot\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}{\sqrt{6}+1}+\frac{2\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}{\sqrt{6}-2}-\frac{4\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[3\cdot\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}+11\right)\left(\sqrt{6}-11\right)=-115\)

\(a,\) ta có : 

\(\Leftrightarrow\left\{{}\begin{matrix}A=\sqrt{3}+\sqrt{2^2.3}-\sqrt{3^2.3}-\sqrt{6^2}\\A=\sqrt{3}+2\sqrt{3}-3\sqrt{3}-6\\A=\sqrt{3}.\left(1+2-3\right)-6\\A=-6\end{matrix}\right.\)

\(\Rightarrow A=-6\) . vậy \(A=9\sqrt{5}\)

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\(b,\) với \(x>0\) và \(x\ne1\) . ta có :

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

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

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

\(\Leftrightarrow B=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(\Leftrightarrow\) \(B=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(\Leftrightarrow B=\dfrac{4}{\sqrt{x}}\)

vậy với \(x>0\) \(;\) \(x\ne1\) thì \(B=\dfrac{4}{\sqrt{x}}\)

để \(B=2\) thì \(\dfrac{4}{\sqrt{x}}=2\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

vậy để \(B=2\) thì \(x=4\)

9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)

10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)

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

11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)

12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)

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

=9-3=6

13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)

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

tìm cả đk giúp mik vs

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

b.

\(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)

\(\Rightarrow A=\dfrac{4+2\sqrt{3}}{\sqrt{3}+1-1}=\dfrac{4+2\sqrt{3}}{\sqrt{3}}=\dfrac{6+4\sqrt{3}}{3}\)

c.

Để \(\sqrt{A}\) xác định \(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)

Ta có:

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

Dấu "=" xảy ra khi \(\sqrt{x}-2=0\Rightarrow x=4\)

a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(=4-\sqrt{15}+\sqrt{15}=4\)

b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

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

\(=2\sqrt{3}\)

c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

\(=2\sqrt{5}+3-2\sqrt{5}+3=6\)

`a, (sqrt 28 - sqrt 12 - sqrt 7) sqrt 7 + 2 sqrt 21`.

`= sqrt(28.7) - sqrt(12.7) - sqrt(7.7) + 2 sqrt 21`.

`= sqrt(4. 7.7) - sqrt (12.7) - 7 + 2 sqrt 21`.

`= 14 - sqrt(4.3.7) - 7 + 2 sqrt 21`.

`= 7`.

`b, (sqrt99-sqrt18-sqrt11)sqrt11+3sqrt22`

`= sqrt(99.11)- sqrt(18.11)-sqrt(11.11) +3sqrt22`

`= sqrt(9.11.11)-sqrt(2.9.11)-11+3sqrt22`

`= 33 - 11 = 22`.

a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\sqrt{2^2\cdot7}-\sqrt{3^2\cdot7}+\dfrac{\sqrt{7}\cdot\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)

\(=2\sqrt{7}-3\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1\)

\(=-\sqrt{7}\)

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

\(=\left[\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

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

\(=\dfrac{2\cdot4}{\sqrt{x}-3}\)

\(=\dfrac{8}{\sqrt{x}-3}\)

b) \(A>B\) khi 

\(\dfrac{8}{\sqrt{x}-3}< -\sqrt{7}\)

\(\Leftrightarrow8< -\sqrt{7x}+3\sqrt{7}\)

\(\Leftrightarrow x< \dfrac{\left(3\sqrt{7}-8\right)^2}{7}\)

a: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

=-2

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

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

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

c: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)

\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)