a) Sửa đề: \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}\)

Ta có: \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}\)

\(=\sqrt{7+2\cdot\sqrt{7}\cdot1+1}-\sqrt{7}\)

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

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

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

=1

b) Ta có: \(B=\sqrt{7+4\sqrt{3}}-2\sqrt{3}\)

\(=\sqrt{4+2\cdot2\cdot\sqrt{3}+3}-2\sqrt{3}\)

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

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

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

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

c) Ta có: \(C=\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}\)

\(=\sqrt{13-2\cdot\sqrt{13}\cdot1+1}+\sqrt{13+2\cdot\sqrt{13}\cdot1+1}\)

\(=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}\)

\(=\left|\sqrt{13}-1\right|+\left|\sqrt{13}+1\right|\)

\(=\sqrt{13}-1+\sqrt{13}+1\)

\(=2\sqrt{13}\)

d) Ta có: \(D=\sqrt{22-2\sqrt{21}}-\sqrt{22+2\sqrt{21}}\)

\(=\sqrt{21-2\cdot\sqrt{21}\cdot1+1}-\sqrt{21+2\cdot\sqrt{21}\cdot1+1}\)

\(=\sqrt{\left(\sqrt{21}-1\right)^2}-\sqrt{\left(\sqrt{21}+1\right)^2}\)

\(=\left|\sqrt{21}-1\right|-\left|\sqrt{21}+1\right|\)

\(=\sqrt{21}-1-\left(\sqrt{21}+1\right)\)

\(=\sqrt{21}-1-\sqrt{21}-1\)

=-2

a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)

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

=7-2

=5

d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)

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

\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)

\(=4\sqrt{7}\)

a, đặt \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)

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

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

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

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

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

\(b,\)

\(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}=\left[\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\right].\sqrt{10-2\sqrt{21}}\)

\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}\right)^2-2\sqrt{7.3}+\left(\sqrt{3}\right)^2}\)

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

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

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

a) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}+\sqrt{2}\right)\)

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

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

=3-1=2

b) Ta có: \(\left(\sqrt{21}+7\right)\cdot\sqrt{10-2\sqrt{21}}\)

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

\(=4\sqrt{7}\)

câu đầu bạn xem lại đề đi nha 

các phần còn lại

b)B=\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)=\(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)=\(\sqrt{7}-1-\left(\sqrt{7}+1\right)=-2\)

c)tính từng căn nha

\(\sqrt{13-4\sqrt{3}}=\sqrt{12-2\sqrt{12}+1}=\sqrt{\left(\sqrt{12}-1\right)^2}=\sqrt{12}-1=2\sqrt{3}-1\)

\(\sqrt{22-12\sqrt{2}}=\sqrt{18-4\sqrt{18}+4}=\sqrt{\left(\sqrt{18}-2\right)^2}=\sqrt{18}-2=3\sqrt{2}-3\)

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

thay vào tính C đc C=2

d)có \(\sqrt{9+4\sqrt{2}}=\sqrt{8+2\sqrt{8}+1}=\sqrt{\left(\sqrt{8}+1\right)^2}=\sqrt{8}+1\)\(\Rightarrow6\sqrt{2+\sqrt{9+4\sqrt{2}}}=6\sqrt{2+\sqrt{8}+1}=6\sqrt{2+2\sqrt{2}+1}\)

=\(6\sqrt{\left(\sqrt{2}+1\right)^2}=6\left(\sqrt{2}+1\right)=6\sqrt{2}+6\)\(\Rightarrow D=\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{17-6\sqrt{2}-6}=\sqrt{11-6\sqrt{2}}=\sqrt{9-6\sqrt{2}+2}\)

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

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

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

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

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

6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)

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

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

a)\(\left(\sqrt{21}+7\right)\cdot\sqrt{10-2\sqrt{21}}\)

\(=\left(\sqrt{21}+7\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)

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

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

b)\(\left(7+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)

\(=\left(7+\sqrt{14}\right)\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)

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

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