Question

Giúp mình với mn ơiii:_)

Answer 1

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

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

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

\(=\dfrac{3\cdot6\sqrt{3}}{\left(2\sqrt{2}\right)^2-\left(3\sqrt{3}\right)^2}\)

\(=\dfrac{18\sqrt{3}}{8-27}\)

\(=-\dfrac{18\sqrt{3}}{19}\)

#Toru

Answer 2

  

Answer 3

\(\dfrac{3}{2\sqrt{2}-3\sqrt{3}}-\dfrac{3}{2\sqrt{2}+3\sqrt{3}}\\ =\dfrac{3\left(2\sqrt{2}+3\sqrt{3}\right)}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}-\dfrac{3\left(2\sqrt{2}-3\sqrt{3}\right)}{\left(2\sqrt{2}+3\sqrt{3}\right)\left(2\sqrt{2}-3\sqrt{3}\right)}\\ =\dfrac{6\sqrt{2}+9\sqrt{3}-6\sqrt{2}+9\sqrt{3}}{8-6\sqrt{6}+6\sqrt{6}-27}\\ =\dfrac{18\sqrt{3}}{-19}\\ =\dfrac{-18\sqrt{3}}{19}\)