Question

Rationalize the denominator 1/3√2-2√3​

Answer 1

Answer:

The answer is $\dfrac{3\sqrt 2+2\sqrt 3}{6}$

Step-by-step explanation:

    $\dfrac{1}{3\sqrt 2-2\sqrt 3}$

$=\dfrac{1}{3\sqrt 2-2\sqrt 3}\times \dfrac{3\sqrt 2+2\sqrt 3}{3\sqrt 2+2\sqrt 3}\\\\=\dfrac{3\sqrt 2+2\sqrt 3}{(3\sqrt 2)^2-(2\sqrt 3)^2}$Using $(a-b)(a+b)=a^2-b^2$

$=\dfrac{3\sqrt 2+2\sqrt 3}{9(2)-4(3)}\\\\=\dfrac{3\sqrt 2+2\sqrt 3}{18-12}=\dfrac{3\sqrt 2+2\sqrt 3}{6}$