Question

[LIÊN]

Trục căn thức ở mẫu

Help me, please 😣😣😣

Answer 1

[Phương An]

\(\dfrac{1+3\sqrt{2}-2\sqrt{3}}{\sqrt{6}+\sqrt{3}+\sqrt{2}}\)

\(=\dfrac{\left[1+\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)\right]\left[\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)\right]}{\left[\sqrt{6}+\left(\sqrt{3}+\sqrt{2}\right)\right]\left[\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)\right]}\)

Tử:

\(\left[1+\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)\right]\left[\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)\right]\)

\(=\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)+6\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{6}\) (nhân phân phối)

\(=5\sqrt{3}-7\sqrt{2}\)

Mẫu:

\(\left[\sqrt{6}+\left(\sqrt{3}+\sqrt{2}\right)\right]\left[\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)\right]\)

\(=6-\left(5+2\sqrt{6}\right)\)

\(=1-2\sqrt{6}\)

Ta có:

\(\dfrac{5\sqrt{3}-7\sqrt{2}}{1-2\sqrt{6}}\)

\(=\dfrac{\left(5\sqrt{3}-7\sqrt{2}\right)\left(1+2\sqrt{6}\right)}{1-24}\)

\(=\dfrac{5\sqrt{3}+30\sqrt{2}-7\sqrt{2}-28\sqrt{3}}{-23}\)

\(=\dfrac{-23\left(\sqrt{3}-\sqrt{2}\right)}{-23}\)

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