Ta có

\(A=\sqrt{11+\sqrt{96}}=\sqrt{11+\sqrt{16.6}}=\sqrt{11+4\sqrt{6}}=\sqrt{8+2.2\sqrt{2}.\sqrt{3}+3}=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}=\left|2\sqrt{2}+\sqrt{3}\right|=2\sqrt{2}+\sqrt{3}=\sqrt{2}+\sqrt{2}+\sqrt{3}\)

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

Ta có \(2>1\Leftrightarrow\sqrt{2}>\sqrt{1}\Leftrightarrow\sqrt{2}>1\Leftrightarrow\sqrt{2}+\sqrt{2}+\sqrt{3}>1+\sqrt{2}+\sqrt{3}\)\(\Leftrightarrow\sqrt{11+\sqrt{96}}>\dfrac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}\)

Vậy A>B

Ta có \(A=\sqrt{11+\sqrt{96}}=\sqrt{11+\sqrt{16.6}}=\sqrt{11+4\sqrt{6}}=\sqrt{8+2.2\sqrt{2}.\sqrt{3}+3}=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}=2\sqrt{2}+\sqrt{3}=\sqrt{3}+\sqrt{2}+\sqrt{2}\)Ta lại có \(B=\dfrac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\dfrac{2\sqrt{2}\left[1-\left(\sqrt{2}-\sqrt{3}\right)\right]}{\left(1+\sqrt{2}-\sqrt{3}\right)\left[1-\left(\sqrt{2}-\sqrt{3}\right)\right]}=\dfrac{2\sqrt{2}\left(1-\sqrt{2}+\sqrt{3}\right)}{1^2-\left(\sqrt{2}-\sqrt{3}\right)^2}=\dfrac{2\sqrt{2}\left(1-\sqrt{2}+\sqrt{3}\right)}{1-\left(2-2\sqrt{6}+3\right)}=\dfrac{2\sqrt{2}\left(1-\sqrt{2}+\sqrt{3}\right)}{1-5+2\sqrt{6}}=\dfrac{2\sqrt{2}\left(1-\sqrt{2}+\sqrt{3}\right)}{2\sqrt{6}-4}=\dfrac{2\sqrt{2}\left(1-\sqrt{2}+\sqrt{3}\right)}{2\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}=\dfrac{1-\sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}}\)\(=\dfrac{\left(1-\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{6}-2+3+\sqrt{6}}{3-2}=\sqrt{3}+\sqrt{2}+1\)Vì \(1< 2\Leftrightarrow\sqrt{1}< \sqrt{2}\Leftrightarrow1< \sqrt{2}\Leftrightarrow\sqrt{3}+\sqrt{2}+1< \sqrt{3}+\sqrt{2}+\sqrt{2}\Leftrightarrow A>B\)

1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)

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

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

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

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

1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)

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

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

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

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

a)

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

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

b)

\(\left(2+\dfrac{11-\sqrt{11}}{1-\sqrt{11}}\right)\left(2+\dfrac{\sqrt{11}+11}{\sqrt{11}+1}\right)\\ =\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{-\left(\sqrt{11}-1\right)}\right)\left(2+\dfrac{\sqrt{11}\left(1+\sqrt{11}\right)}{\sqrt{11}+1}\right)\\ =\left(2-\sqrt{11}\right)\left(2+\sqrt{11}\right)\\ =4-11\\ =-7\)

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

=(căn 3-căn 2)(căn 3+căn 2)

=3-2=1

b: \(=\left(2-\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{\sqrt{11}-1}\right)\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}+1\right)}{\sqrt{11}+1}\right)\)

=(2-căn 11)(2+căn 11)

=4-11

=-7

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

\(=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}\cdot1+1^2}+\left|\sqrt{2}-2\right|\)

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

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

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

\(=3\)

b) \(\dfrac{1}{5}\sqrt{50}-2\sqrt{96}-\dfrac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\dfrac{1}{6}}\)

\(=\dfrac{1}{5}\cdot5\sqrt{2}-2\cdot4\sqrt{6}-\sqrt{\dfrac{30}{15}}+\sqrt{\dfrac{144}{6}}\)

\(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}\)

\(=-8\sqrt{6}+2\sqrt{6}\)

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

c) \(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)

\(=\left[\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}-2\right]\left[\dfrac{4\left(1-\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+4\right]\)

\(=\left(\sqrt{5}-1-2\right)\left(\dfrac{4\left(1-\sqrt{5}\right)}{1-5}+4\right)\)

\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}-1+4\right)\)

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

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

\(=-4\)

a) \(\sqrt[]{3+2\sqrt[]{2}}+\sqrt[]{\left(\sqrt[]{2}-2\right)^2}\)

\(=\sqrt[]{2+2\sqrt[]{2}.1+1}+\left|\sqrt[]{2}-2\right|\)

\(=\sqrt[]{\left(\sqrt[]{2}+1\right)^2}+\left(2-\sqrt[]{2}\right)\) \(\left(\left(\sqrt[]{2}\right)^2=2< 2^2=4\right)\)

\(=\left|\sqrt[]{2}+1\right|+2-\sqrt[]{2}\)

\(=\sqrt[]{2}+1+2-\sqrt[]{2}\)

\(=3\)

a: Ta có: \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

b: Ta có: \(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)

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

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

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

a,ĐKXĐ: tự tìm :v

 \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(x+2\sqrt{x}+1\right)-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+1\right)^2-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6+2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{9\sqrt{x}-x-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

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

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

\(=\dfrac{\left(\sqrt{x}-1\right)\left(10-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\dfrac{10-\sqrt{x}}{\sqrt{x}+3}\)

mk chỉ bt làm phần b thôi.

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