Question

rationalize the denominator 1÷√3-√2

Answer 1

hope this helps........

Answer 2

ANSWER:

To Rationalize:

• 1/(√3-√2)

Solution:

$\text{We are given that,}\\\\:\longrightarrow\dfrac{1}{\sqrt3-\sqrt2}\\\\\text{So,}\\\\:\implies\dfrac{1}{\sqrt3-\sqrt2}\\\\:\implies\dfrac{1}{\sqrt3-\sqrt2}\times\dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\\\\:\implies\dfrac{\sqrt3+\sqrt2}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\\\\\text{We know that,}\\\\:\hookrightarrow (a+b)(a-b)=a^2-b^2\\\\\text{So,}\\\\:\implies\dfrac{\sqrt3+\sqrt2}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\\\\:\implies\dfrac{\sqrt3+\sqrt2}{(\sqrt3)^2-(\sqrt2)^2}\\\\:\implies\dfrac{\sqrt3+\sqrt2}{3-2}\\\\:\implies\dfrac{\sqrt3+\sqrt2}{1}\\\\\bf{:\implies\sqrt3+\sqrt2}$

Formulae Used:

• (a - b)(a + b) = a² - b²

Learn More:

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identities}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\bf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\bf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\bf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) - B^{3}\\\\8)\bf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\9)\bf\: A^{3} - B^{3} = (A-B)(A^{2} + AB + B^{2})\\\\ \end{minipage}}$