Question

solve this problem please rationalised the denominator ​

Answer 1

Answer:

hopevit help u

Step-by-step explanation:

1st)2.5

2nd)3.2is 2.3

Answer 2

Solution!!

(1)

$\sf \dfrac{5-3\sqrt{14}}{7+2\sqrt{14}}$

$\sf =\dfrac{5-3\sqrt{14}}{7+2\sqrt{14}}\times \dfrac{7-2\sqrt{14}}{7-2\sqrt{14}}$

$\sf =\dfrac{(5-3\sqrt{14})(7-2\sqrt{14})}{(7+2\sqrt{14})(7-2\sqrt{14})}$

$\sf =\dfrac{35-10\sqrt{14}-21\sqrt{14}+84}{49-4\times 14}$

$\sf =\dfrac{119-31\sqrt{14}}{49-56}$

$\sf =\dfrac{119-31\sqrt{14}}{-7}$

$\sf =-\dfrac{119-31\sqrt{14}}{7}$

(2)

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

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

$</p><p>[tex]\sf =\dfrac{(2+3\sqrt{3})(\sqrt{2}+\sqrt{3})}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}$

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

$\sf =\dfrac{(2+3\sqrt{3})(\sqrt{2}+\sqrt{3})}{-1}$

$\sf =-(2+3\sqrt{3})(\sqrt{2}+\sqrt{3})$

$\sf =-2\sqrt{2}-2\sqrt{3}-3\sqrt{6}-9$