Question

rationalise the denominator of: 1/roo3 + root2 - root1​

Answer 1

Step-by-step explanation:

Given √3+1/√3-1

We need to rationalize the givenequation

To rationalize we will multiply and divide the denominator by √3+1

(√3+1)√3+1)/(√3-1)(√3+1)

Denominator can be simplified using the identity (a2-b2)=(a-b)(a+b)

And the equation becomes

=(√3+1)2/(√3)2-(1)2

=(4+2√3)/2

=2+√3

Answer 2

Answer:

$Multiply numerator and denominator by [√3+(√2+1)] ,we get</p><p></p><p>=\frac{\sqrt{3}+\sqrt{2}+1}{[\sqrt{3}-(\sqrt{2}+1)][\sqrt{3}+(\sqrt{2}+1)]}=[3−(2+1)][3+(2+1)]3+2+1</p><p></p><p>/* By algebraic identity:</p><p></p><p>\boxed {(a-b)(a+b)=a^{2}-b^{2}}(a−b)(a+b)=a2−b2 */</p><p></p><p>=\frac{\sqrt{3}+\sqrt{2}+1}{(\sqrt{3})^{2}-(\sqrt{2}+1)^{2}}=(3)2−(2+1)23+2+1</p><p></p><p>=\frac{\sqrt{3}+\sqrt{2}+1}{3-(2+1+2\sqrt{2})}=3−(2+1+22)3+2+1</p><p></p><p>=\frac{\sqrt{3}+\sqrt{2}+1}{3-(3+2\sqrt{2})}=3−(3+22)3+2+1</p><p></p><p>=\frac{\sqrt{3}+\sqrt{2}+1}{3-3-2\sqrt{2}}=3−3−223+2+1</p><p></p><p>=\frac{\sqrt{3}+\sqrt{2}+1}{-2\sqrt{2}}=−223+2+1</p><p></p><p>Multiply numerator and denominator by √2 , we get</p><p></p><p>=\frac{\sqrt{2}(\sqrt{3}+\sqrt{2}+1)}{-2\sqrt{2}\times \sqrt{2}}=−22×22(3+2+1)</p><p></p><p>=\frac{\sqrt{6}+2+\sqrt{2}}{-4}=−46+2+2</p><p></p><p>Therefore,</p><p></p><p>\frac{1}{\sqrt{3}-\sqrt{2}-1} < /p > < p > =\frac{\sqrt{6}+2+\sqrt{2}}{-4}3−2−11</p><p>=−46+2+2</p><p></p><p>•••♪</p><p></p><p>$