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(i) Simplify :- (13 + 23 + 33)1/2
(ii)
Simplify :- \left ( \frac{3}{5} \right )^{4}\left ( \frac{8}{5} \right )^{-12}\left ( \frac{32}{5} \right )^{6}
(iii)
Simplify :- \left ( \frac{1}{27} \right )^{-\frac{2}{3}}
(iv) Simplify :- \left [ \left \{ \left ( 625 \right )^{-\frac{1}{2}} \right \}^{-\frac{1}{4}} \right ]^{2}
(v) Simplify :- \frac{9^{\frac{1}{3}}\times 27^{\frac{1}{2}}}{3^{\frac{1}{6}}\times 9^{-\frac{2}{3}}}
(vi) Simplify :-64^{-\frac{1}{3}}\left ( 64^{\frac{1}{3}}-64^{\frac{2}{3}} \right )
(vii) Simplify :- \frac{8^{\frac{1}{3}}\times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}


 

(i) Answer.          6
Solution.     (13 + 23 + 33)1/2
We know that
13 = 1.1.1 = 1
23 = 2.2.2 = 8
33 = 3.3.3 = 27
Putting these values we get
(13 + 23 + 33)1/2 = \sqrt{1+8+27}
=\sqrt{36}= 6
Hence the answer is 6

(ii) Answer.  \frac{2025}{64}       

Solution.\left ( \frac{3}{5} \right )^{4}\left ( \frac{8}{5} \right )^{-12}\left ( \frac{32}{5} \right )^{6}
 We know that

8 = 2.2.2 = 23
32 = 2.2.2.2.2 = 25
\left ( \frac{3}{5} \right )^{4}\left ( \frac{8}{5} \right )^{-12}\left ( \frac{32}{5} \right )^{6}= \left ( \frac{3}{5} \right )^{4}\left ( \frac{2^{3}}{5} \right )^{-12}\left ( \frac{2^{5}}{5} \right )^{6}
= \frac{3^{4}\left ( 2^{3} \right )^{-12}\left ( 2^{5} \right )^{6}}{5^{4}5^{-12}5^{6}}                \because \left ( \frac{a}{b} \right )^{m}= \frac{a^{m}}{b^{m}}
= \frac{3^{4}2^{-36}2^{30}}{5^{4}5^{-12}5^{6}}                            \because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}
= \frac{3^{4}\times 2^{-36+30}}{5^{4-12+6}}                        \because \left ( a \right )^{m}\left ( a \right )^{n}= \left ( a \right )^{m+n}
= \frac{3^{4}\times 2^{-6}}{5^{-2}}
= \frac{3^{4}\times 5^{2}}{2^{6}}                                   \because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}
= \frac{81\times 25}{64}
= \frac{2025}{64}
Hence the answer is \frac{2025}{64}
     
(iii) Answer. 9
Solution. Given \left ( \frac{1}{27} \right )^{-\frac{2}{3}}
We know that
27 = 3.3.3 = 33
\left ( \frac{1}{27} \right )^{-\frac{2}{3}}= \left ( \frac{1}{3^{3}} \right )^{-\frac{2}{3}}
= \left ( 3^{3} \right )^{\frac{2}{3}}            \because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}
= \left ( 3 \right )^{3\times \frac{2}{3}}         \because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}

= 32 = 9
Hence the answer is 9

(iv) Answer. 5
Solution. Given \left [ \left \{ \left ( 625 \right )^{-\frac{1}{2}} \right \}^{-\frac{1}{4}} \right ]^{2}
We know that
625= \left ( 25 \right )\left ( 25 \right )= 5\cdot 5\cdot 5\cdot 5= 5^{4}
\left [ \left \{ \left ( 625 \right )^{-\frac{1}{2}} \right \}^{-\frac{1}{4}} \right ]^{2}= \left [ \left \{ \left ( \left ( 5 \right )^{4} \right )^{-\frac{1}{2}} \right \} ^{-\frac{1}{4}}\right ]^{2}
= 5^{4\times \frac{-1}{2}\times \frac{-1}{4}\times 2}        \because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}

= 51 = 5
Hence the answer is 5

(v) Answer.      \sqrt[3]{\frac{1}{3}}
Solution. We have \frac{9^{\frac{1}{3}}\times 27^{\frac{1}{2}}}{3^{\frac{1}{6}}\times 9^{-\frac{2}{3}}}
Now we know that
9 = 3.3 = 32
27 = 3.3.3 = 33
\frac{9^{\frac{1}{3}}\times 27^{\frac{1}{2}}}{3^{\frac{1}{6}}\times 9^{-\frac{2}{3}}}= \frac{\left ( 3^{2} \right )^{\frac{1}{3}}\times \left ( 3^{3} \right )^{\frac{1}{2}}}{\left ( 3 \right )^{\frac{1}{6}}\times \left ( 3^{2} \right )^{\tfrac{-2}{3}}}
= \frac{\left ( 3\right )^{2\times \frac{1}{3}}\times \left ( 3 \right )^{3\times \frac{-1}{2}}}{\left ( 3 \right )^{\frac{1}{6}}\times \left ( 3 \right )^{\tfrac{-2}{3}}}        \because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}
= \frac{\left ( 3 \right )^{\frac{2}{3}-\frac{3}{2}}}{\left ( 3 \right )^{\frac{1}{6}-\frac{2}{3}}}              \because \left ( a \right )^{m}\left ( a \right )^{n}= \left ( a \right )^{m+n}
= \frac{3^{\frac{4-9}{6}}}{3^{\frac{1-4}{6}}}
=\frac{3^{-\frac{5}{6}}}{3^{-\frac{3}{6}}}
= 3^{\frac{-5}{6}-\left ( \frac{-3}{6} \right )}           \because \frac{\left ( a \right )^{m}}{\left ( a \right )^{n}}= \left ( a \right )^{m-n}
= 3^{-\frac{2}{6}}
= \left ( \frac{1}{3} \right )^{\frac{1}{3}}                 \because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}
= \sqrt[3]{\frac{1}{3}}

Hence the answer is \sqrt[3]{\frac{1}{3}}

(vi) Answer. – 3
Solution. We have ,64^{-\frac{1}{3}}\left ( 64^{\frac{1}{3}}-64^{\frac{2}{3}} \right )
We know that 64 =4.4.4=43
= \left ( 4^{3} \right )^{\frac{-1}{3}}\left \{ \left ( \left ( 4^{3} \right ) ^{\frac{1}{3}}-\left ( 4^{3} \right ) ^{\frac{2}{3}}\right )\right \}
= \left ( 4 \right )^{3\times \frac{-1}{3}}\left \{ \left ( \left ( 4 \right )^{3\times \frac{1}{3}}-\left ( 4 \right )^{3\times \frac{2}{3}} \right ) \right \}      \because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}
= 4^{-1}\left ( 4-4^{2} \right )
= \frac{1}{4}\left ( 4-16 \right )                                                        \because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}
= \frac{1}{4}\left ( -12 \right )

= – 3
Hence the answer is – 3
 

(vii) Answer. 16
Solution.
Given ,\frac{8^{\frac{1}{3}}\times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}
We know that
8 = 2.2.2 = 23
16 = 2.2.2.2 = 24
32 = 2.2.2.2.2 = 25
\frac{8^{\frac{1}{3}}\times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}= \frac{\left ( 2^{3} \right )^{\frac{1}{3}}\times\left ( 2^{4} \right )^{\frac{1}{3}}}{\left ( 2^{5} \right )^{-\frac{1}{3}}}
= \frac{2^{3\times\frac{1}{3}}\times2^{4\times\frac{1}{3}}}{2^{5\times\frac{-1}{3}}}              \because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}       
= 2^{1+\frac{4}{3}+\frac{5}{3}}                             \because \left ( a \right )^{m}\left ( a \right )^{n}= \left ( a \right )^{m+n}  and
                                                       \because \frac{\left ( a \right )^{m}}{\left ( a \right )^{n}}= \left ( a \right )^{m-n}
= 2^{\frac{3+4+5}{3}}= 2^{\frac{12}{3}}
= 2^{4}= 16

Hence the answer is 16.

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(i) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{4}{\sqrt{3}}
(ii) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{6}{\sqrt{6}}
(iii) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}
(iv) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{2}{2+\sqrt{2}}
(v) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{1}{\sqrt{3}+\sqrt{2}}

(i) Answer.   2.3093
Solution. Given: \frac{4}{\sqrt{3}}
Rationalising,
\frac{4}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}= \frac{4\sqrt{3}}{3}
(Given that \sqrt{3}= 1.732)
= \frac{4\times 1\cdot 732}{3}
= 2.3093
Hence the answer is 2.3093

(ii) Answer.  2.449
Solution.  Given: \frac{6}{\sqrt{6}}
Rationalising,
\frac{6}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}= \frac{6\sqrt{6}}{6}
= \frac{6\times \sqrt{2}\sqrt{3}}{6}
Putting the given values,
\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732
We get :
= \sqrt{2}\cdot \sqrt{3}
= 1\cdot 414\times 1\cdot 732= 2\cdot 449
Hence the answer is 2.449

(iii) Answer.  0.462852
Solution.   Given that \frac{\sqrt{10}-\sqrt{5}}{2}
This can be written as

\frac{\sqrt{2}\times \sqrt{5}-\sqrt{5}}{2}
Now putting the given values,

\sqrt{2}= 1\cdot 414,\sqrt{5}= 2\cdot 236
We get :
\Rightarrow \frac{1\cdot 414\times 2\cdot 236-2\cdot 236}{2}
= 0.462852
  Hence the answer is 0.462852

(iv) Answer.  0.414
Solution.  Given: \frac{\sqrt{2}}{2+\sqrt{2}}
Rationalising,
\frac{\sqrt{2}}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}
Using   (a – b) (a + b) = a2 – b2
= \frac{\sqrt{2}\left ( 2-\sqrt{2} \right )}{2^{2}-\sqrt{2}^{2}}
= \frac{2\sqrt{2}-2}{4-2}
= \frac{2\left ( \sqrt{2}-1 \right )}{2}
= \sqrt{2}-1

Putting the given value of \sqrt{2}= 1\cdot 414
We get
= 1.414 – 1
= 0.414
 Hence the answer is 0.414

(v)  Answer.  0.318
Solution. Given that \frac{1}{\sqrt{3}+\sqrt{2}}
Rationalising,
\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}
Using   (a – b) (a + b) = a2 – b2
= \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}^{2}-\sqrt{2}^{2}}
= \frac{\sqrt{3}-\sqrt{2}}{3-2}
= \sqrt{3}-\sqrt{2}

Putting the given values, \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732
We get,
= 1.732 – 1.414
= 0.318
Hence the answer is 0.318

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infoexpert27

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(i) Find the values of a in each of the following : \frac{5+2\sqrt{3}}{7+4\sqrt{3}}= a-6\sqrt{3}
(ii) Find the values of a in the following : \frac{3-\sqrt{5}}{3+2\sqrt{5}}= a\sqrt{5}-\frac{19}{11}

(iii) Find the values of b in the following : \frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}
(iv) Find the values of a and b in the following : \frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}= a+\frac{7}{11}\sqrt{5b}

(i) Answer.   a = 11
Solution.   We have, \frac{5+2\sqrt{3}}{7+4\sqrt{3}}= a-6\sqrt{3}
LHS = \frac{5+2\sqrt{3}}{7+4\sqrt{3}}
Rationalising the denominator, we get:
= \frac{5+2\sqrt{3}}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}
= \frac{\left ( 5+2\sqrt{3} \right )\left ( 7-4\sqrt{3} \right )}{\left ( 7+4\sqrt{3} \right )\left ( 7-4\sqrt{3} \right )}
{Using (a – b) (a + b) = a2 – b2}

= \frac{35+14\sqrt{3}-20\sqrt{3}-24}{7^{2}-\left ( 4\sqrt{3} \right )^{2}}
= \frac{11-6\sqrt{3}}{49-48}
= 11-6\sqrt{3}
Now RHS = a-6\sqrt{3}
\Rightarrow 11-6\sqrt{3}= a-6\sqrt{3}
\Rightarrow 11-6\sqrt{3}= a-6\sqrt{3}
\Rightarrow a= 11
Hence a = 11 is the required answer

(ii)Answer.   a= \frac{9}{11}
Solution.  Given that, \frac{3-\sqrt{5}}{3+2\sqrt{5}}= a\sqrt{5}-\frac{19}{11}

LHS = \frac{3-\sqrt{5}}{3+2\sqrt{5}}
Rationalising the denominator, we get:
LHS = \frac{3-\sqrt{5}}{3+2\sqrt{5}}\times \frac{3-2\sqrt{5}}{3-2\sqrt{5}}
{Using (a – b) (a + b) = a2 – b2}
= \frac{9-3\sqrt{5}-6\sqrt{5}+10}{3^{2}-\left ( 2\sqrt{5} \right )^{2}}
= \frac{19-9\sqrt{5}}{9-20}
Now RHS = a\sqrt{5}-\frac{19}{11}
\Rightarrow \frac{9\sqrt{5}}{11}-\frac{19}{11}= a\sqrt{5}-\frac{19}{11}
Comparing both , we get
\Rightarrow a= \frac{9}{11}
Hence a= \frac{9}{11}is the correct answer

(iii) Answer:b = -\frac{5 }{6}

Solution:

Given:

\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}

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

Rationalize

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

=\frac{\sqrt{2}(3 \sqrt{2}+2 \sqrt{3})+\sqrt{3}(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2})^{2}-(2 \sqrt{3})^{2}}

= \frac{6+2 \sqrt{6}+3 \sqrt{6}+6}{18-12}

= 2+\frac{5 \sqrt{6}}{6}

2+\frac{5 \sqrt{6}}{6}=2-b \sqrt{6}

b = -\frac{5 }{6}

 

(iv) Answer. a = 0, b = 1
Solution.         Given,
\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}= a+\frac{7}{11}\sqrt{5b}
LHS = \frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}
=\frac{\left ( 7+\sqrt{5} \right )\times\left ( 7+\sqrt{5} \right )-\left ( 7-\sqrt{5} \right ) \times \left ( 7-\sqrt{5} \right )}{\left (7 -\sqrt{5} \right )\left (7 +\sqrt{5} \right )}
Using (a – b) (a + b) = a2 – b2
(a + b)2 = a2 + b2 + 2ab
(a - b)2 = a2 + b2 - 2ab
= \frac{\left ( 7^{2} +\sqrt{5}^{2}+2\cdot 7\cdot \sqrt{5}\right )-\left ( 7^{2} +\sqrt{5}^{2}-2\cdot 7\cdot \sqrt{5} \right )}{7^{2}-\sqrt{5}^{2}}
= \frac{\left ( 49+5+14\sqrt{5} \right )-\left ( 49+5-14\sqrt{5} \right )}{49-5}
= \frac{54+14\sqrt{5}-54+14\sqrt{5}}{44}
= \frac{28\sqrt{5}}{44}
RHS = a+\frac{7}{11}\sqrt{5b}
Now LHS = RHS
\Rightarrow \frac{28\sqrt{5}}{44}= a+\frac{7}{11}\sqrt{5b}
\Rightarrow 0+\left ( \frac{4}{4} \right )\frac{7}{11}\sqrt{5}= a+\frac{7}{11}\sqrt{5b}

\Rightarrow a = 0, b = 1
Hence the answer is a = 0, b = 1

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(i) Rationalise the denominator of the following : \frac{2}{3\sqrt{3}}
(ii)Rationalise the denominator of the following : \frac{\sqrt{40}}{\sqrt{3}}
(iii) Rationalise the denominator of the following : \frac{3+\sqrt{2}}{4\sqrt{2}}
(iv)Rationalise the denominator of the following :\frac{16}{\sqrt{41}-5}
(v) Rationalise the denominator of the following : \frac{2+\sqrt{3}}{2-\sqrt{3}}
(vi) Rationalise the denominator of the following  : \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}
(vii) Rationalise the denominator of the following : \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}

(viii) Rationalise the denominator of the following : \frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}

(ix) Rationalise the denominator of the following : \frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}


 

(i) Answer.  \frac{2\sqrt{3}}{9}  
Solution.         We have, \frac{2}{3\sqrt{3}}
Rationalising the denominator, we get:
\frac{2}{3\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}
= \frac{2\sqrt{3}}{3\sqrt{3}\sqrt{3}}
= \frac{2\sqrt{3}}{9}
Hence the answer is \frac{2\sqrt{3}}{9}
 

(ii) Answer.  \frac{2\sqrt{30}}{3}  
Solution. We have ,\frac{\sqrt{40}}{\sqrt{3}}
We know that, 40 = (2) (2) (10)
\frac{\sqrt{40}}{\sqrt{3}}= \frac{\sqrt{2\cdot 2\cdot 10}}{\sqrt{3}}= \frac{2\sqrt{10}}{\sqrt{3}}

Rationalising the denominator, we get:
= \frac{2\sqrt{10}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}
= \frac{2\sqrt{10}\sqrt{3}}{\sqrt{3}\sqrt{3}}
= \frac{2\sqrt{30}}{3}

Hence the answer is: \frac{2\sqrt{30}}{3}

(iii) Answer.  \frac{3\sqrt{2}+2}{8}
Solution.  We have \frac{3+\sqrt{2}}{4\sqrt{2}}
Rationalising the denominator, we get:
\frac{3+\sqrt{2}}{4\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}
= \frac{\left ( 3+\sqrt{2} \right )\sqrt{2}}{4\sqrt{2}\sqrt{2}}
= \frac{3\sqrt{2}+2}{8}
Hence the answer is \frac{3\sqrt{2}+2}{8}

(iv) Answer. \sqrt{41}+5
Solution. We have \frac{16}{\sqrt{41}-5}
Rationalising the denominator, we get:
\frac{16}{\sqrt{41}-5}\times \frac{\sqrt{41}+5}{\sqrt{41}+5}
= \frac{16\left ( \sqrt{41}+5 \right )}{\left ( \sqrt{41}-5 \right )\sqrt{41}+5}
Using  the identity (a – b) (a + b) = a2 – b2
We get:
= \frac{16\left ( \sqrt{41}+5 \right )}{\left ( \sqrt{41} \right )^{2}-\left ( 5 \right )^{2}}
= \frac{16\left ( \sqrt{41}+5 \right )}{41-25}
= \frac{16\left ( \sqrt{41}+5 \right )}{16}
= \sqrt{41}+5
Hence the answer is \sqrt{41}+5

(v) Answer.    7+4\sqrt{3}
Solution. We have, \frac{2+\sqrt{3}}{2-\sqrt{3}}
Rationalising the denominator, we get:
\frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2-\sqrt{3}}
= \frac{\left ( 2+\sqrt{3} \right )^{2}}{\left ( 2-\sqrt{3} \right )\left ( 2+\sqrt{3} \right )}

Using   (a – b) (a + b) = a2 – b2
and      (a + b)2 = a2 + b2 + 2ab
= \frac{2^{2}+\left ( \sqrt{3} \right )^{2}+2\cdot 2\cdot \sqrt{3}}{2^{2}-\left ( \sqrt{3} \right )^{2}}
= \frac{4+3+4\sqrt{3}}{4-3}
= 7+4\sqrt{3}
Hence the answer is 7+4\sqrt{3}

(vi)Answer.  3\sqrt{2}-2\sqrt{3}
Solution.   We have, \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}
Rationalising the denominator, we get:
= \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}
= \frac{\sqrt{6}\left ( \sqrt{3}-\sqrt{2} \right )}{\left ( \sqrt{2}+\sqrt{3} \right )\left ( \sqrt{3}-\sqrt{2} \right )}
Using  the identity (a – b) (a + b) = a2 – b2
We get:
= \frac{\sqrt{18}-\sqrt{12}}{\left ( \sqrt{3} \right )^{2}\left ( \sqrt{2} \right )^{2}}
= \frac{\sqrt{3\cdot 3\cdot 2}-\sqrt{2\cdot 2\cdot 3}}{3-2}
= \frac{3\sqrt{2}-2\sqrt{3}}{1}
= 3\sqrt{2}-2\sqrt{3}
Hence the answer 3\sqrt{2}-2\sqrt{3}

(vii) Answer. 5+2\sqrt{6}
Solution. We have, \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}

Rationalising the denominator, we get:
\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}
= \frac{\left ( \sqrt{3} +\sqrt{2}\right )^{2}}{\left ( \sqrt{3} -\sqrt{2} \right )\left ( \sqrt{3} +\sqrt{2} \right )}

Using   (a – b) (a + b) = a2 – b2
and      (a + b)2 = a2 + b2 + 2ab
= \frac{\left ( \sqrt{3} \right )^{2}+\left ( \sqrt{2} \right )^{2}+2\sqrt{3}\sqrt{2}}{\left ( \sqrt{3} \right )^{2}-\left ( \sqrt{2} \right )^{2}}
= \frac{3+2+2\sqrt{6}}{3-2}
= 5+2\sqrt{6}
Hence the answer is 5+2\sqrt{6}

(viii)

Answer: 9+2 \sqrt{15}

Solution:

We have \frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}

Rationalize

=\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{3 \sqrt{5}(\sqrt{5}+\sqrt{3})+\sqrt{3}(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}

=\frac{15+3 \sqrt{15}+\sqrt{15}+3}{5-3}=\frac{18+4 \sqrt{15}}{2}

=9+2 \sqrt{15}

(ix) Answer: \frac{9+4 \sqrt{6}}{15}

Solution:

We have \frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}

=\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}=\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{16 \times 3}+\sqrt{9 \times 2}}=\frac{4 \sqrt{3}+5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}}

Rationalize

=\frac{4 \sqrt{3}+5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}} \times \frac{(4 \sqrt{3}-3 \sqrt{2})}{(4 \sqrt{3}-3 \sqrt{2})}

=\frac{4 \sqrt{3}(4 \sqrt{3}-3 \sqrt{2})+5 \sqrt{2}(4 \sqrt{3}-3 \sqrt{2})}{(4 \sqrt{3})^{2}-(3 \sqrt{2})^{2}}

=\frac{48-12 \sqrt{6}+20 \sqrt{6}-30}{30}

=\frac{18+8 \sqrt{6}}{30}

=\frac{9+4 \sqrt{6}}{15}

 

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infoexpert27

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(i) Simplify the following : \sqrt{45}-3\sqrt{20}+4\sqrt{5}
(ii) Simplify the following : \frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}
(iii) Simplify the following : 4\sqrt{12}\times 7\sqrt{6}
(iv) Simplify the following : 4\sqrt{28}\div 3\sqrt{7}\div 3\sqrt{7}
(v) Simplify the following : 3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}
(vi) Simplify the following : \left ( \sqrt{3}-\sqrt{2} \right )^{2}
(vii) Simplify the following : \sqrt[4]{81}- 8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{255}
(viii) Simplify the following : \frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}
(ix) Simplify the following : \frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}






 

(i) Answer. \sqrt{5}
Solution.    \sqrt{45}-3\sqrt{20}+4\sqrt{5} 
We know that,
45 = 3\times 3\times 5
20 = 2\times 2\times 5
So we get
\sqrt{3\times3\times5 }-3\sqrt{2\times2\times5}+4\sqrt{5}
= 3\sqrt{5}-3\left ( 2\sqrt{5} \right )+4\sqrt{5}
= 3\sqrt{5}-6\sqrt{5}+4\sqrt{5}
= 7\sqrt{5}-6\sqrt{5}
= \sqrt{5}  
Hence the answer is \sqrt{5}

(ii)  Answer. \frac{7\sqrt{6}}{12}
Solution. We have, \frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}
We know that,

24= 6\times 4= 3\times 2\times 2\times 2
54= 9\times 6= 3\times 3\times 3\times 2
So we get
\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}= \frac{2\sqrt{6}}{8}+\frac{3\sqrt{6}}{9}
= \frac{\sqrt{6}}{4}+\frac{\sqrt{6}}{3}
Taking LCM (3,4) = 12

= \frac{3\sqrt{6}+4\sqrt{6}}{12}
= \frac{7\sqrt{6}}{12}

(iii) Answer.  \sqrt[28]{2^{18} \times 3^{11}}
Solution.   We have
\sqrt[4]{12}\times \sqrt[7]{6}
We know that
12 = 2\times 2\times 3
6 = 2\times 3
So we get,
=\sqrt[4]{2\times 2\times 3}\times \sqrt[7]{2\times 3}  
=2^{1 / 4} \cdot 2^{1 / 4} \cdot 3^{1 / 4} \cdot 2^{1 / 7} \cdot 3^{1 / 7}
=2^{\frac{1}{4}+\frac{1}{4}+\frac{1}{7}} \times 3^{\frac{1}{4}+\frac{1}{7}}

=2^{9 / 14} \times 3^{11 / 28}

=\sqrt[28]{2^{18} \times 3^{11}}
Hence the number is \sqrt[28]{2^{18} \times 3^{11}}.

(iv)Answer.    \frac{8}{\left ( 3\times \sqrt[3]{7} \right )}     
Solution.   We have, 4\sqrt{28}\div 3\sqrt{7}\div \sqrt[3]{7}
We know that
28 = 4\times 7
So we can write,
4\sqrt{28}\div 3\sqrt{7}\div \sqrt[3]{7}= \left [ \frac{4\sqrt{28}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}
= \left [ \frac{4\sqrt{4\times 7}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}

= \left [ \frac{4\times 2\sqrt{ 7}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}

= \frac{8}{3}\div 7^{\frac{1}{3}}
= \frac{8}{\left ( 3\times 7^{\frac{1}{3}} \right )}
= \frac{8}{\left ( 3\times \sqrt[3]{7} \right )}
  Hence the answer is \frac{8}{\left ( 3\times \sqrt[3]{7} \right )}

(v) Answer.  3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}
We know that
27 = 3\times 3\times 3
So, 3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}} = 3\sqrt{3}+2\sqrt{3\times 3\times 3}+\frac{7}{\sqrt{3}}

= 3\sqrt{3}+2\left ( 3\sqrt{3} \right )+\frac{7}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}                
(Rationalising the denominator)

= 3\sqrt{3}+6\left ( \sqrt{3} \right )+\frac{7\sqrt{3}}{3}

= \left ( 3+6+\frac{7}{3} \right )\sqrt{3}         (Taking \sqrt{3} common)
Now LCM (1,1,3) = 3
= \left ( \frac{9+18+7}{3} \right )\sqrt{3}
= \frac{34}{3}\sqrt{3}
= 19\cdot 63

Hence the answer is 19.63

(vi) Answer. 5-2\sqrt{6}
Solution. Given, \left ( \sqrt{3}-\sqrt{2} \right )^{2}
We know that (a + b)2 = a2 – 2ab + b2
Comparing the given equation with the identity, we get: