KnowledgeBoat Logo
|

Mathematics

Rationalize the denominators of the following:

(i) 17\dfrac{1}{\sqrt{7}}

(ii) 17−6\dfrac{1}{\sqrt{7} - \sqrt{6} }

(iii)15+2\dfrac{1}{\sqrt{5} + \sqrt{2} }

(iv)17−2\dfrac{1}{\sqrt{7} - 2 }

Number System

5 Likes

Answer

(i)17\dfrac{1}{\sqrt{7}}

Multiply by (7)(\sqrt{7}) in numerator and denominator

= 17\dfrac{1}{\sqrt{7}} x 77\dfrac{\sqrt{7}}{\sqrt{7}}

= 77\dfrac{\sqrt{7}}{7}

Hence, 17\dfrac{1}{\sqrt{7}} = 77\dfrac{\sqrt{7}}{7}

(ii) 17−6\dfrac{1}{\sqrt{7} - \sqrt{6}}

Multiply by 17+6\dfrac{1}{\sqrt{7} + \sqrt{6}} in numerator and denominator

= 17−6\dfrac{1}{\sqrt{7} - \sqrt{6} } x 7+67+6\dfrac{\sqrt{7}+ \sqrt{6}}{\sqrt{7} + \sqrt{6}}

= 7+6(7)2−(6)2\dfrac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (6)^2 }

= 7+67−6\dfrac{\sqrt{7} + \sqrt{6} }{7 - 6}

= 7+61\dfrac{\sqrt{7} + \sqrt{6} }{1}

= (7+6)(\sqrt{7} + \sqrt{6})

=Hence, 17−6\dfrac{1}{\sqrt{7} - \sqrt{6}} = (7+6)(\sqrt{7} + \sqrt{6})

(iii)15+2\dfrac{1}{\sqrt{5} + \sqrt{2}}

Multiply by 5+2{\sqrt{5} + \sqrt{2}} in numerator and denominator

= 15+2\dfrac{1}{\sqrt{5} + \sqrt{2}} x 5−25−2\dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}

= 5−2(5)2−(2)2\dfrac{\sqrt{5} - \sqrt{2}} {(\sqrt{5})^2 - (\sqrt{2})^2 }

= 5−25−2\dfrac{\sqrt{5} - \sqrt{2}} {5 - 2}

= 5−23\dfrac{\sqrt{5} - \sqrt{2}} {3}

=Hence, 15+2\dfrac{1}{\sqrt{5} + \sqrt{2}} = 5−23\dfrac{\sqrt{5} - \sqrt{2}} {3}

(iv)17−2\dfrac{1}{\sqrt{7} - 2}

Multiply by 7+2{\sqrt{7} + {2}} in numerator and denominator

= 17−2\dfrac{1}{\sqrt{7} - 2} X 7+27+2\dfrac{\sqrt{7} + 2 }{\sqrt{7} + 2 }

= 7+2(7)2−(2)2\dfrac{\sqrt{7} + 2} {(\sqrt{7})^2 - (2)^2 }

= 7+27−4\dfrac{\sqrt{7} + 2}{7 - 4}

= 7+23\dfrac{\sqrt{7} + 2}{3}

= Hence, 17−2\dfrac{1}{\sqrt{7} - 2 } = 7+23\dfrac{\sqrt{7} + 2}{3}

Answered By

2 Likes

Related Questions