Xem lại câu c) và d) 

b: =căn 10-3+4-căn 10=1

a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)

câu a coi lai có sai sót j ko

\(A=\sqrt{11-2\sqrt{10}}+\sqrt{9-2\sqrt{4}}-\sqrt{10}-\sqrt{7}\)

\(=\sqrt{\left(\sqrt{10}-1\right)^2}+\sqrt{5}-\sqrt{10}-\sqrt{7}=\sqrt{10}-1+\sqrt{5}-\sqrt{10}-\sqrt{7}\)

\(=\sqrt{5}-\sqrt{7}-1\)

\(=\left|3-\sqrt{10}\right|+\sqrt{\left(\sqrt{10}-1\right)^2}\\ =\sqrt{10}-3+\sqrt{10}-1=\sqrt{10}-4\)

Mấy bạn giải dùm mình nha thanks 

\(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2-2\sqrt{2.3}+3}+\sqrt{2+2\sqrt{2.3}+3}=\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}=2\sqrt{3}\)

\(\sqrt{7-2\sqrt{10}}+\sqrt{7+2\sqrt{10}}=\sqrt{2-2\sqrt{2.5}+5}+\sqrt{2+2\sqrt{2.5}+5}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}=2\sqrt{5}\)

\(\sqrt{11+2\sqrt{8}}+\sqrt{11-2\sqrt{8}}\)

Ta có: \(\left(\sqrt{11+2\sqrt{8}}+\sqrt{11-2\sqrt{8}}\right)^2=11+2\sqrt{8}+11-2\sqrt{8}+2\sqrt{\left(11+2\sqrt{8}\right)\left(11-2\sqrt{8}\right)}=22+2\sqrt{121-32}=22+2\sqrt{89}\)

=>\(\sqrt{11+2\sqrt{8}}+\sqrt{11-2\sqrt{8}}=\sqrt{22+2\sqrt{89}}\)

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

b) \(\sqrt{7-2\sqrt{10}}+\sqrt{7+2\sqrt{10}}=\left(\sqrt{5}-\sqrt{2}\right)+\left(\sqrt{5}+\sqrt{2}\right)=2\sqrt{5}\)

c) \(\sqrt{11+2\sqrt{8}}+\sqrt{11-2\sqrt{8}}=chả-biết-nữa\)

sorry

\(\sqrt{7-\sqrt{24}}-\dfrac{\sqrt{50}-5}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11+\sqrt{120}\right)\left(11+2\sqrt{30}\right)^2}\)

\(=\sqrt{7-2\sqrt{6}}-\dfrac{5\left(\sqrt{2}-1\right)}{\sqrt{5}\left(\sqrt{2}-1\right)}+\left|11+2\sqrt{30}\right|\sqrt{11-2\sqrt{30}}\)

\(=\sqrt{1^2-2\sqrt{6}\cdot1+\left(\sqrt{6}\right)^2}-\dfrac{\sqrt{5}\cdot\sqrt{5}}{\sqrt{5}}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{5}\cdot\sqrt{6}+\left(\sqrt{5}\right)^2}\)

\(=\sqrt{\left(1-\sqrt{6}\right)^2}-\sqrt{5}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\)

\(=\left|1-\sqrt{6}\right|-\sqrt{5}+\left(11+2\sqrt{30}\right)\left|\sqrt{6}-\sqrt{5}\right|\)

\(=-1+6-\sqrt{5}+\left(\sqrt{6}+\sqrt{5}\right)^2\left(\sqrt{6}-\sqrt{5}\right)\)

\(=\sqrt{6}-1-\sqrt{5}+\left[\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2\right]\left(\sqrt{6}+\sqrt{5}\right)\)

\(=\sqrt{6}-1-\sqrt{5}+\left(6-5\right)\left(\sqrt{6}+\sqrt{5}\right)\)

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

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

\(=\sqrt{6+1-2\sqrt{6}}-\dfrac{\sqrt{5}\left(\sqrt{10}-\sqrt{5}\right)}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11-\sqrt{120}\right)\left(11+\sqrt{120}\right)^2}\\ =\sqrt{\left(\sqrt{6}-\sqrt{1}\right)^2}-\sqrt{5}+\sqrt{\left(11^2-120\right)\left(11+2\sqrt{30}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{1\left(6+5+2\sqrt{6\cdot5}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{6}+\sqrt{5}=2\sqrt{6}-\sqrt{1}\)

\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)

a) \(\sqrt{11+2\sqrt[]{18}}\)

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

\(=\sqrt{9+2.3\sqrt[]{2}+2}\)

\(=\sqrt{\left(3+\sqrt[]{2}\right)^2}=\left|3+\sqrt[]{2}\right|=3+\sqrt[]{2}\)

b) \(\sqrt[]{7+2\sqrt[]{10}}\)

\(=\sqrt[]{7+2\sqrt[]{5}.\sqrt[]{2}}\)

\(=\sqrt[]{5+2\sqrt[]{5}.\sqrt[]{2}+2}\)

\(=\sqrt[]{\left(\sqrt[]{5}+\sqrt[]{2}\right)^2}=\left|\sqrt[]{5}+\sqrt[]{2}\right|=\sqrt[]{5}+\sqrt[]{2}\)

c) \(\sqrt[]{7+4\sqrt[]{3}}\)

\(=\sqrt[]{4+2.2\sqrt[]{3}+3}\)

\(=\sqrt[]{\left(2+\sqrt[]{3}\right)^2}=\left|2+\sqrt[]{3}\right|=2+\sqrt[]{3}\)

d) \(\sqrt[]{16-2\sqrt[]{55}}\) \(\left(12\rightarrow16\right)\)

\(=\sqrt[]{11-2\sqrt[]{5}.\sqrt[]{11}+5}\)

\(=\sqrt[]{\left(\sqrt[]{11}-\sqrt[]{5}\right)^2}==\left|\sqrt[]{11}-\sqrt[]{5}\right|=\sqrt[]{11}-\sqrt[]{5}\left(\sqrt[]{11}>\sqrt[]{5}\right)\)