a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)

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

c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}=\sqrt{\left(11-6\sqrt{2}\right)\left(11+6\sqrt{2}\right)}=\sqrt{121-72}=\sqrt{49}=7\)

a: Ta có: \(\sqrt{75}-2\sqrt{27}+\sqrt{48}\)

\(=5\sqrt{3}-2\cdot3\sqrt{3}+4\sqrt{3}\)

\(=3\sqrt{3}\)

c: Ta có: \(\sqrt{8+2\sqrt{7}}-\sqrt{11-4\sqrt{7}}\)

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

=3

\(\sqrt{3\cdot27}-\sqrt{\dfrac{144}{36}}\)=\(\sqrt{81}-\sqrt{4}\)=9-2=7

\(\dfrac{2\cdot3+3\cdot6}{4}\)=6

\(\sqrt{7}-\sqrt{7-2\cdot\sqrt{7}+1}\)=\(\sqrt{7}-\left(\sqrt{7}-1\right)\)=1

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

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

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

\(\sqrt{3}\cdot\sqrt{9}+5\cdot\sqrt{4}\cdot3-2\sqrt{3}\)

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

=\(11\sqrt{3}\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\\ =\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\\ =\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\\ =\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)=6-121=-115\)

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

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

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

\(=6-121=-115\)

Các bạn có thể giúp mình đc ko ? Mình cảm ơn rất nhiều !

\(VT=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)

\(=4+\sqrt{11}-3\sqrt{7}\)

a: Ta có: \(A=\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)

\(=\left(2+\sqrt{2}\right):\left(2+\sqrt{2}\right)\)

=1

b: Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)

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

=1

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

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

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

\(=\dfrac{12\left(3-\sqrt{3}\right)}{6}-2\sqrt{3}+\dfrac{3\sqrt{2}}{2}-\sqrt{3}\)

\(=2\left(3-\sqrt{3}\right)-3\sqrt{3}+\dfrac{3\sqrt{2}}{2}=6-5\sqrt{3}+\dfrac{3\sqrt{2}}{2}\) (câu này khả năng đề sai, dấu \(\sqrt{3}.\sqrt{2}\) ở mẫu cuối cùng là dấu trừ mới hợp lý)

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

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

Dấu giữa 2 dấu ngoặc là dấu chia sẽ hợp lý hơn