If a = sqrt{21} - sqrt{20} and b = sqrt{18} - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We are given $a = \sqrt{21} - \sqrt{20}$ and $b = \sqrt{18} - \sqrt{17}$.To compare $a$ and $b$,we rationalize the expressions:$a = \frac{(\sqrt{2

Vedclass · Accepted Answer

$a < b$

Vedclass · Answer

(D) We are given $a = \sqrt{21} - \sqrt{20}$ and $b = \sqrt{18} - \sqrt{17}$.To compare $a$ and $b$,we rationalize the expressions:$a = \frac{(\sqrt{21} - \sqrt{20})(\sqrt{21} + \sqrt{20})}{\sqrt{21} + \sqrt{20}} = \frac{21 - 20}{\sqrt{21} + \sqrt{20}} = \frac{1}{\sqrt{21} + \sqrt{20}}$.Similarly,$b = \frac{(\sqrt{18} - \sqrt{17})(\sqrt{18} + \sqrt{17})}{\sqrt{18} + \sqrt{17}} = \frac{18 - 17}{\sqrt{18} + \sqrt{17}} = \frac{1}{\sqrt{18} + \sqrt{17}}$.Since $\sqrt{21} + \sqrt{20} > \sqrt{18} + \sqrt{17}$,it follows that $\frac{1}{\sqrt{21} + \sqrt{20}} Therefore,$a < b$.

If $a = \sqrt{21} - \sqrt{20}$ and $b = \sqrt{18} - \sqrt{17}$,then

Similar Questions

$\frac{\sqrt{5/2} + \sqrt{7 - 3\sqrt{5}}}{\sqrt{7/2} + \sqrt{16 - 5\sqrt{7}}} = $

The solution of the equation $4 \cdot 9^{x - 1} = 3 \cdot \sqrt{2^{2x + 1}}$ is

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

Evaluate: $\sqrt{3 + \sqrt{5}}$

If $3^x - 3^{x - 1} = 6$,then $x^x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module