Question

rationalise
1+√2/2-√2​

Answer 1

Solution!!

$\sf \dfrac{1+\sqrt{2}}{2-\sqrt{2}}$

Multiply the fraction by $\sf \dfrac{2+\sqrt{2}}{2+\sqrt{2}}$.

$\sf =\dfrac{1+\sqrt{2}}{2-\sqrt{2}}\times \dfrac{2+\sqrt{2}}{2+\sqrt{2}}$

To multiply the fractions, multiply the numerator and denominator separately.

$\sf =\dfrac{(1+\sqrt{2})(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})}$

Multiply the parentheses.

$\sf =\dfrac{2+\sqrt{2}+2\sqrt{2}+2}{(2-\sqrt{2})(2+\sqrt{2})}$

Use (a - b)(a + b) = a² - b² to simplify the product in the denominator.

$\sf =\dfrac{2+\sqrt{2}+2\sqrt{2}+2}{(2)^{2}-(\sqrt{2})^{2}}$

$\sf =\dfrac{2+\sqrt{2}+2\sqrt{2}+2}{4-2}$

Add the numbers in the numerator.

$\sf =\dfrac{4+\sqrt{2}+2\sqrt{2}}{4-2}$

Collect the like terms in the numerator.

$\sf =\dfrac{4+3\sqrt{2}}{4-2}$

Subtract the numbers in the denominator.

$\sf =\dfrac{4+3\sqrt{2}}{2}$