Question

Thu gọn:

\(A=\frac{\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}-\sqrt{3-2\sqrt{2}}\)

có thể  bị sai đề ( sai thì các bạn giúp mình sửa lại nhé )

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