Question

$\huge{ {\blue{\fbox\red{ǫᴜᴇsᴛɪᴏɴ}}}}$

Rationalize the denominator of:
$\frac{1}{ \sqrt{3} - \sqrt{2} + 1 }$
​

Answer 1

$\Huge\bf\maltese{\underline{\green{Answe°᭄}}}\maltese$

$\implies$$\large\bf{\underline{\red{VERIFIED✔}}}$

$</p><p>\begin{gathered}Given \: \\\frac{1}{\sqrt{3}-\sqrt{2}-1}\end{gathered}Given3−2−11 \\ \\ </p><p>Multiply numerator and denominator by [√3+(√2+1)] ,we get \\ \\ </p><p>=\frac{\sqrt{3}+\sqrt{2}+1}{[\sqrt{3}-(\sqrt{2}+1)][\sqrt{3}+(\sqrt{2}+1)]}=[3−(2+1)][3+(2+1)]3+2+1</p><p>/ \\ \\ * By algebraic identity:</p><p>\boxed {(a-b)(a+b)=a^{2}-b^{2}}(a−b)(a+b)=a2−b2 */ \\ \\ </p><p>=\frac{\sqrt{3}+\sqrt{2}+1}{(\sqrt{3})^{2}-(\sqrt{2}+1)^{2}} \\ \\ =(3)2−(2+1)23+2+1</p><p> \\ \\ =\frac{\sqrt{3}+\sqrt{2}+1}{3-(2+1+2\sqrt{2})} \\ \\ =3−(2+1+22)3+2+1</p><p> \\ \\ =\frac{\sqrt{3}+\sqrt{2}+1}{3-(3+2\sqrt{2})} \\ \\ =3−(3+22)3+2+1</p><p> \\ \\ =\frac{\sqrt{3}+\sqrt{2}+1}{3-3-2\sqrt{2}}=3−3−223+2+1</p><p> \\ \\ =\frac{\sqrt{3}+\sqrt{2}+1}{-2\sqrt{2}} \\ \\ =−223+2+1</p><p>Multiply numerator and denominator by √2 , we get</p><p> \\ \\ =\frac{\sqrt{2}(\sqrt{3}+\sqrt{2}+1)}{-2\sqrt{2}\times \sqrt{2}} \\ \\ =−22×22(3+2+1)</p><p> \\ \\ =\frac{\sqrt{6}+2+\sqrt{2}}{-4}=−46+2+2</p><p> \\ \\ Therefore, \\ \\ </p><p>\frac{1}{\sqrt{3}-\sqrt{2}-1} < /p > < p > \\ \\ =\frac{\sqrt{6}+2+\sqrt{2}}{-4}3</p><p>$

• I Hope it's Helpful My Friend Surbhisingh.

Answer 2

$\huge\bold{\textbf{\textsf{{\color{cyan}{Answer}}}}}$

$</p><p>Given3−2−11Given3−2−11</p><p>Multiply numerator and denominator by [√3+(√2+1)] ,we get</p><p>=[3−(2+1)][3+(2+1)]3+2+1=[3−(2+1)][3+(2+1)]3+2+1</p><p>/∗ By algebraic identity:</p><p>(a−b)(a+b)=a2−b2(a−b)(a+b)=a2−b2 ∗/</p><p>=(3)2−(2+1)23+2+1=(3)2−(2+1)23+2+1</p><p>=3−(2+1+22)3+2+1=3−(2+1+22)3+2+1</p><p>=3−(3+22)3+2+1=3−(3+22)3+2+1</p><p>=3−3−223+2+1=3−3−223+2+1</p><p>=−223+2+1=−223+2+1</p><p>Multiply numerator and denominator by √2 , we get</p><p>=−22×22(3+2+1)=−22×22(3+2+1)</p><p>=−46</p><p>$

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