The number obtained on rationalizing the denominator of $\frac{1}{\sqrt{7} - 2}$ is
$\frac{\sqrt{7} + 2}{45}$
$\frac{\sqrt{7} + 2}{5}$
$\frac{\sqrt{7} + 2}{3}$
$\frac{\sqrt{7} - 2}{3}$

Question

The number obtained on rationalizing the denominator of dfrac -1-sqrt -7- - 2- isdfrac -sqrt -7- - 2-45-dfrac -sqrt -7- - 2-5-dfrac -sqrt -7- - 2-3-dfrac -sqrt -7- - 2-3-

Toppr Admin · Accepted Answer

We use the identity -a-x221A-b-a-x2212-x221A-b-a2-x2212-b-Multiply and divide 1-x221A-7-x2212-2 by -x221A-7-2-xA0-to get-xA0-1-x221A-7-x2212-2-xD7-x221A-7-2-x221A-7-2-x221A-7-27-x2212-4-x221A-7-23-xA0-Hence- option A-

The number obtained on rationalizing the denominator of 1√7−2 is√7+245√7+25√7+23√7−23

We use the identity (a+√b)(a−√b)=a2−b.Multiply and divide 1√7−2 by √7+2 to get 1√7−2×√7+2√7+2=√7+27−4=√7+23 Hence, option A.

The number obtained on rationalizing the denominator of $\frac{1}{\sqrt{7} - 2}$ is
$\frac{\sqrt{7} + 2}{45}$
$\frac{\sqrt{7} + 2}{5}$
$\frac{\sqrt{7} + 2}{3}$
$\frac{\sqrt{7} - 2}{3}$

We use the identity $(a + \sqrt{b}) (a - \sqrt{b}) = a^{2} - b$ .
Multiply and divide $\frac{1}{\sqrt{7} - 2}$ by $\sqrt{7} + 2$ to get
$\frac{1}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{\sqrt{7} + 2}{7 - 4} = \frac{\sqrt{7} + 2}{3}$

Hence, option $A .$