Rationalize the denominators of the following- i 1 3- 2 ii 1 - 7 - 6 iii 1 - 7

(i) 13+√(2)=13+√(2)×3 √(2)3 √(2)=3 √(2)(3)^2 (√(2))^2=3 √(2)7 (ii) 1√(7) √(6)=1√(7) √(6)×√(7)+√(6)√(7)+√(6)=√(7)+√(6)(√(7))^2 (√(6))^2=√(7)+√(6) (iii) ...

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Standard IX

Mathematics

Rationalisation

Rationalize t...

Question

Rationalize the denominators of the following:
i) $\frac{1}{3 + \sqrt{2}}$
ii) $\frac{1}{\sqrt{7} - \sqrt{6}}$
iii) $\frac{1}{\sqrt{7}}$

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Solution

(i) $\frac{1}{3 + \sqrt{2}} = \frac{1}{3 + \sqrt{2}} \times \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = \frac{3 - \sqrt{2}}{(3)^{2} - (\sqrt{2})^{2}} = \frac{3 - \sqrt{2}}{7}$

(ii) $\frac{1}{\sqrt{7} - \sqrt{6}} = \frac{1}{\sqrt{7} - \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^{2} - (\sqrt{6})^{2}} = \sqrt{7} + \sqrt{6}$

(iii) $\frac{1}{\sqrt{7}} = \frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7}$

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Similar questions

Q. Simplify each of the following by rationalizing the denominator:
i)

\frac{6 - 4 \sqrt{2}}{6 + 4 \sqrt{2}}

ii)

\frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} + \sqrt{5}}

iii)

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

Rationalize the denominator of the following:

$\frac{1}{(\sqrt{7} - \sqrt{6})}$

Rationalize the denominator

$\frac{1}{(\sqrt{7} - \sqrt{6})}$

Rationalise the denominators of the following:

(i) (ii)

(iii) (iv)

Question 5 (ii)
Rationalize the denominator of :
$\frac{1}{(\sqrt{7} - \sqrt{6})}$ .

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Rationalize the denominators of the following: i) 13+√2ii) 1√7−√6iii) 1√7

(i)13+√2=13+√2×3−√23−√2=3−√2(3)2−(√2)2=3−√27(ii)1√7−√6=1√7−√6×√7+√6√7+√6=√7+√6(√7)2−(√6)2=√7+√6(iii)1√7=1√7×√7√7=√77

Rationalize the denominators of the following:
i) $\frac{1}{3 + \sqrt{2}}$
ii) $\frac{1}{\sqrt{7} - \sqrt{6}}$
iii) $\frac{1}{\sqrt{7}}$