Question

rationalize the denominator ​

Answer 1

Answer:

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According to question:-

$\frac{ \sqrt[7]{3} }{ \sqrt{10} + \sqrt{3} } - \frac{ \sqrt[2]{5} }{ \sqrt{6} + \sqrt{5} } - \frac{ \sqrt[3]{2} }{ \sqrt{15} + \sqrt[3]{2} }$

Firstly, rationalize the denominator of

$\frac{ \sqrt[7]{3} }{ \sqrt{10} + \sqrt{3} }$

$\frac{ \sqrt[7]{3} }{ \sqrt{10} + \sqrt{3} } \times \frac{ \sqrt{10} - \sqrt{3} }{ \sqrt{10} - \sqrt{3} }$

$\frac{ \sqrt[7]{30} - \sqrt[7]{9} }{( { \sqrt{10} })^{2} - ( { \sqrt{3} })^{2} }$

$= \frac{ \sqrt[7]{30} - \sqrt[7]{9} }{10 - 3}$

$= \frac{7( \sqrt{30} - \sqrt{9}) }{7}$

$= \sqrt{30} - \sqrt{9} ......(1)$

Now, rationalize the denominator of:-

$\frac{ \sqrt[2]{5} }{ \sqrt{6} + \sqrt{5} }$

$\frac{ \sqrt[2]{5} }{ \sqrt{6} + \sqrt{5} } \times \frac{ \sqrt{6} - \sqrt{5} }{ \sqrt{6} - \sqrt{5} }$

$\frac{ \sqrt[2]{30} - \sqrt[2]{25} }{( { \sqrt{6} })^{2} - ( { \sqrt{5} })^{2} }$

$= \frac{ \sqrt[2]{30} - \sqrt[2]{25} }{6 - 5}$

$= 2( \sqrt{30} - \sqrt{25} ......(2)$

Now, rationalize the denominator of:-

$\frac{ \sqrt[3]{2} }{ \sqrt{15} + \sqrt[3]{2} }$

$\frac{ \sqrt[3]{2} }{ \sqrt{15} + \sqrt[3]{2} } \times \frac{ \sqrt{15} - \sqrt[3]{2} }{ \sqrt{15} - \sqrt[3]{2} }$

$\frac{ \sqrt[3]{30} - \sqrt[9]{4} }{( { \sqrt{15} })^{2} - ( { \sqrt[3]{2} })^{2} }$

$= \frac{ \sqrt[3]{30} - \sqrt[9]{4} }{15 - 18}$

$= \frac{3( \sqrt{30} - \sqrt[3]{4}) }{ - 3}$

$= - ( \sqrt{30} - \sqrt[3]{4} ) \: or \: - \sqrt{30} + \sqrt[3]{4} .......(3)$

According to question:-

=$( \sqrt{30} - \sqrt{9} ) - (2( \sqrt{30} - \sqrt{25} ) - ( - \sqrt{30} + \sqrt[3]{4}$

=$\sqrt{30} - \sqrt{9} - \sqrt[2]{30} + \sqrt[2]{25} + \sqrt{30} - \sqrt[3]{4}$

Cancelling √30+√30 with -2√30:

$= - \sqrt{9} + \sqrt[2]{25} - \sqrt[3]{4}$

√9 will become 3, √25 will become 5 and √4 will become 2

$- 3 + 2 \times 5 - 3 \times 2$

$- 3 + 10 - 6$

$= 10 - 9$

$= 1$Answer:-

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Answer 2

Step-by-step explanation:

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