a) S=\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}.\)
\(\sqrt{2}.\)S=\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}.\)
\(\sqrt{2}.\)S =\(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(\sqrt{2}.\)S=|\(\sqrt{7}\)-1|+|\(\sqrt{7}\)+1|=\(\sqrt{7}\)-1-\(\sqrt{7}\)-1=- 2
S= - \(\sqrt{2}.\)
b)\(=\dfrac{\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}=...\)
b/ \(A=\dfrac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
\(A=\dfrac{\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(A=\dfrac{3}{1}=3\)
a) \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(\Rightarrow A^2=\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2=4-\sqrt{7}-2\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}+4+\sqrt{7}\)=\(8-2\sqrt{16-7}=8-2\sqrt{9}=8-6=2\)
=> A =\(-\sqrt{2}\)
A= căn 2 loại vì A lớn hơn 0
\(B=\dfrac{\sqrt{3}+\sqrt{9+6\sqrt{2}+2}-\sqrt{2+2\sqrt{6}+3}}{\sqrt{2}+\sqrt{5+2\sqrt{5+1}-\sqrt{5+2\sqrt{10}+2}}}\)
theo các hằng đẳng thức mà làm ra nhá , kq = 3 , mới tính nhẩm , mk ko làm ra vì giờ bận rồi