Math, asked by guri477, 1 month ago

Rationalise the denominator
(i) 1/√7 (ii) 1/√7-√6
(iii) 1/√5+√2 (iv) 1/√7-2​

Answers

Answered by rahulkumarr179
5

Answer:

1)√7/7

2)(√7+√6)

3)(√5-√2)/3

4)(√7+2)/3

hope it helps u dear☺️☺️☺️

Answered by MoodyCloud
36

(i)  \sf \dfrac{1}{\sqrt{7}}

 \longrightarrow \sf \dfrac{1}{\sqrt{7}} \times \dfrac{\sqrt{7}}{\sqrt{7}}

 \longrightarrow \pmb{\sf \dfrac{\sqrt{7}}{7}}

Hence, Rationalised!!

(ii)  \sf \dfrac{1}{\sqrt{7} - \sqrt{6}}

 \longrightarrow \sf \dfrac{1}{\sqrt{7}-\sqrt{6}} \times \dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}

 \longrightarrow \sf \dfrac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}

 \longrightarrow \sf \dfrac{\sqrt{7}+\sqrt{6}}{7-6}

 \longrightarrow \sf \dfrac{\sqrt{7}+\sqrt{6}}{1}

 \longrightarrow \pmb{\sf \sqrt{7}+\sqrt{6}}

Hence, Rationalised!!

(iii)  \sf \dfrac{1}{\sqrt{5} + \sqrt{2}}

 \longrightarrow \sf \dfrac{1}{\sqrt{5} + \sqrt{2}} \times \dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5} - \sqrt{3}}

 \longrightarrow \sf \dfrac{\sqrt{5 }- \sqrt{2}}{(\sqrt{5})^{2}- (\sqrt{2})^{2}}

 \longrightarrow \sf \dfrac{\sqrt{5}- \sqrt{2}}{5-2}

 \longrightarrow \pmb{\sf \dfrac{\sqrt{5}-\sqrt{2}}{3}}

Hence, Rationalised!!

(iv)  \sf \dfrac{1}{\sqrt{7} - 2}

 \longrightarrow \sf \dfrac{1}{\sqrt{7} - 2} \times \dfrac{\sqrt{7} + 2}{\sqrt{7}+2}

 \longrightarrow \sf \dfrac{\sqrt{7} + 2}{(\sqrt{7})^{2}-(2)^{2}}

 \longrightarrow \sf \dfrac{\sqrt{7} + 2}{7-4}

 \longrightarrow \pmb{\sf \dfrac{\sqrt{7} + 2}{3}}

Hence, Rationalised!!

Identity used :-

(a + b) (a - b) = a² - b²

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