Question

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

Answer 1

bdfbzdtvbeay           q4etwrtc3t5wtư

Answer 2

êrtechcgrgdcgtgư

Answer 3

nhân cả tử và mẫu với \(\sqrt{2}\)nha

Answer 4

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

\(=\frac{\sqrt{2-2\cdot1\sqrt{2}+1}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{2+2\cdot1\sqrt{2}+1}}{\sqrt{7+12\sqrt{2}}}.\)

\(=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{9+12\sqrt{2}+8}}.\)

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

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

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

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

\(=\frac{8-6}{9-8}=\frac{2}{1}=2\)