Question

Rút gọn biểu thức

\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{17+12\sqrt{2}}\)

Answer 1

Phải là \(\sqrt{17+12\sqrt{2}}\) chớ bạn :<

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

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

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

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

\(=2\)