Đặt:   \(A=\frac{\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}\)\(>\)\(0\)

=>   \(A^2=\frac{7+\sqrt{5}+2.\sqrt{\left(7+\sqrt{5}\right)\left(7-\sqrt{5}\right)}+7-\sqrt{5}}{7+2\sqrt{11}}\)

            \(=\frac{14+4\sqrt{11}}{7+2\sqrt{11}}\)

             \(=\frac{2\left(7+2\sqrt{11}\right)}{7+2\sqrt{11}}=2\)

=>  \(A=\sqrt{2}\)

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

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

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

Đặt y= \(\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}\)

=> y² = \(\left(\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}\right)^2\)= \(\left(\sqrt{7+\sqrt{5}}\right)^2+2\sqrt{\left(7+\sqrt{5}\right)\left(7-\sqrt{5}\right)}+\left(\sqrt{7-\sqrt{5}}\right)^2\)

=\(7+\sqrt{5}+2\sqrt{7^2-\left(\sqrt{5}\right)^2}+7-\sqrt{5}\)= \(14+2\sqrt{44}\)= \(14+4\sqrt{11}\)= \(2\left(7+2\sqrt{11}\right)\)

=> y= \(\sqrt{2\left(7+2\sqrt{11}\right)}\)

=> A = \(\frac{\sqrt{2\left(7+2\sqrt{11}\right)}}{\sqrt{7+2\sqrt{11}}}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-\left|\sqrt{2}-1\right|=\sqrt{2}-\left(\sqrt{2}-1\right)\left(do\sqrt{2}>1\right)=\sqrt{2}-\sqrt{2}+1=0+1=1\)

= 1 nha ban

=1 nha

Đặt:    \(B=\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}\)

=>    \(B^2=7+\sqrt{5}+7-\sqrt{5}+2\sqrt{\left(7+\sqrt{5}\right)\left(7-\sqrt{5}\right)}\)

=>   \(B^2=14+2\sqrt{49-5}\)

=>   \(B^2=14+2\sqrt{44}\)

=>   \(A=\frac{\sqrt{14+4\sqrt{11}}}{7+2\sqrt{11}}-\sqrt{\left(\sqrt{2}-1\right)^2}\)

=>   \(A=\sqrt{\frac{2}{7+2\sqrt{11}}}-\left(\sqrt{2}-1\right)\)

=>   \(A=\sqrt{\frac{2}{7+2\sqrt{11}}}-\sqrt{2}+1\)

ĐỀ BÀI CHẮC SAI RỒI PHẢI DƯỚI MẪU PHẢI LÀ    \(\sqrt{7+2\sqrt{11}}\)    THÌ LÚC ĐÓ BIỂU THỨC A RA ĐẸP HƠN !!!!

NẾU SỬA ĐỀ BÀI NHƯ TRÊN:

=>    \(A=\frac{\sqrt{2}.\sqrt{7+2\sqrt{11}}}{\sqrt{7+2\sqrt{11}}}-\left(\sqrt{2}-1\right)\)

=>   \(A=\sqrt{2}-\sqrt{2}+1\)

=>   \(A=1\)

ĐÓ BÂY GIỜ RA A  = 1 RẤT ĐẸP

\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=\frac{32-6-2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}=\frac{14-\sqrt{5}}{2}\) \(\left(\frac{9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2-\left(\frac{9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2=\left(\frac{9-2\sqrt{14}-9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)\left(\frac{9-2\sqrt{14}+9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)=\frac{-72\sqrt{14}}{\sqrt{7}-\sqrt{2}}\)

câu a coi lai có sai sót j ko

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

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

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

a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)

b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)