Question

5. Rationalise the denominator.
1 / (√3) + (√2-1)​

Answer 1

Answer:

Step-by-step explanation:

= 1/√3 - √2 - 1 × √3 + √2 + 1

= √3 + √2 + 1/√3 + √2 + 1

= √3 + √2 + 1/(√3)^2 - (√2)^2 -(1)^2

= √3 + √2 + 1/3-2-1

= √3 + √2 + 1/0

= √3 + √2 + 1 is ur answer.

Step-by-step explanation:

hope u have been understood thanku

Answer 2

Step-by-step explanation:

$\frac{1}{\sqrt{3} +\sqrt{2} -1}$

keeping $\sqrt{3}+\sqrt{2}$ as "A" and $1$ as "B"

USING, (A-B)(A+B)=$A^{2} - B^{2}$

$\frac{1}{\sqrt{3} +\sqrt{2} -1}$   X   $\frac{\sqrt{3} +\sqrt{2}+1} \sqrt{3} +\sqrt{2}+1$

$\frac{\sqrt{3}+\sqrt{2} }{(\sqrt{3} +\sqrt{2} )^{2} -(1)^{2} }$

$\frac{\sqrt{3}+\sqrt{2} }{[(\sqrt{3} )^{2}+2X\sqrt{6} +(\sqrt{2} ^{2})]-1 }$ =$\frac{\sqrt{3} +\sqrt{2} }{3+2\sqrt{6} +2-1}$=$\frac{\sqrt{3}+\sqrt{2} }{4+2\sqrt{6} }$

We need to re rationalize it.

$\frac{\sqrt{3}+\sqrt{2} }{4+2\sqrt{6} } X\frac{4-2\sqrt{6} }{4-2\sqrt{6} }$ =$\frac{4\sqrt{3}-2\sqrt{18} +4\sqrt{2}-2\sqrt{12} }{(4)^{2}-(2\sqrt{6})^{2} }$=$\frac{4\sqrt{3}-2\sqrt{18} +4\sqrt{2}-2\sqrt{12} }{16-144 }$=$\frac{4\sqrt{3}-2\sqrt{18} +4\sqrt{2}-2\sqrt{12} }{-128 }$=

$\frac{-4\sqrt{3}+2\sqrt{18} -4\sqrt{2}+2\sqrt{12} }{128 }$

This hand typed by me..hope this helps you...if it did mark me as the brainliest...