bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không

Chia nhỏ ra bạn ơi!

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

\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)

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

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

\(A=4-5\)

\(A=-1\)

____

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

\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)

\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)

\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

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

\(B=6-121\)

\(B=-115\)

a: Ta có: \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)

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

=-5+2

=-3

d: \(=\sqrt{5}\left(\sqrt{3}-1\right)-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)

=căn 5-1/2*căn 5

=1/2*căn 5

e: \(=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{2}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}=\dfrac{1}{\sqrt{6}}\)

f:=2+căn 3+căn 2-2-căn 3=căn 2

4: Ta có: \(\dfrac{1}{3+\sqrt{5}}-\dfrac{1}{3-\sqrt{5}}\)

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

\(=\dfrac{-\sqrt{5}}{2}\)

Bài 1: 

\(D=\dfrac{1}{2}\sqrt{48}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}.4\sqrt{3}-\sqrt{3}+5.\dfrac{2\sqrt{3}}{3}=2\sqrt{3}-\sqrt{3}+\dfrac{10\sqrt{3}}{3}=\dfrac{3\sqrt{3}+10\sqrt{3}}{3}=\dfrac{13\sqrt{3}}{3}\)

\(E=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}-\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{9-5}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{9-5}}=\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}=-\sqrt{5}\)

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

Bài 2: 

Ta có: G-1

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

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

\(=\dfrac{-\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}\le0\forall x\) thỏa mãn ĐKXĐ

hay \(G\le1\)

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

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

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

\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{2}\right)\)

\(=\left[-\dfrac{\sqrt{2}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}-\sqrt{5}\right]\left(\sqrt{5}-\sqrt{2}\right)\)

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

\(=-3\)

\(\dfrac{2}{7+4\sqrt{3}}+\dfrac{2}{7-4\sqrt{3}}\)

\(=\dfrac{2\left(7-4\sqrt{3}\right)+2\left(7+4\sqrt{3}\right)}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

\(=\dfrac{14-8\sqrt{3}+14+8\sqrt{3}}{49-48}\)

= 28

\(\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{2}{3-\sqrt{5}}}\)

\(=\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{4}{6-2\sqrt{5}}}\)

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

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

\(=\dfrac{2\sqrt{5}-2-2\sqrt{5}-2}{5-1}\)

= - 1

\(\dfrac{4}{1-\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\)

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

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

\(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)

\(=\dfrac{2}{4+\sqrt{6+2\sqrt{5}}}\) (nhân [căn 2] vào cả tử và mẫu)

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

\(=\dfrac{2}{5+\sqrt{5}}=\dfrac{2\left(5-\sqrt{5}\right)}{25-5}=\dfrac{5-\sqrt{5}}{10}\)