Question

please solve these sums​

Answer 1

Answer:

Step-by-step explanation:

Here is your answer

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Answer 2

Hi there, :)

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1) ( 3 - √3)²

Ans :

Use

[( a - b )² = a² +b² - 2ab]

using formula;

( 3 - √3)² = 3² + (√3)²- 2x3x√3

              = 9 + 3 - 6√3

              =  12 - 6√3

further simplifying , (by taking common)

=  6 ( 2 - √3) [Ans.]

2) (√5 - √2) ( √2 - √3 )  (we can say it is in the simplest form already) (or)

Ans:

we multiply directly ,

i.e;  

(√5x√2 - √5x√3 - √2x√2 + √2x√3)

we get

( √10 - √15 + √6 -2)(cannot be further simplified)

=( √10 - √15 + √6 -2) [ Ans.]

3) (5 + √7)²

Ans:

Using expansion:

[(a + b)² = a² + b² + 2ab]

we get

(5 + √7)² = 25 + 7 + 10√7

              = 32 + 10√7

              = 2( 16 + 5 √7)       (simplest form) [Ans.]

4) $\frac{1}{\sqrt{3} + \sqrt{2} } + \frac{2}{\sqrt{5} - \sqrt{3} } - \frac{3}{\sqrt{2} + \sqrt{5} }$

Now we rationalize each term of the equation :  

(i.e multiply and divide with its negative counterpart)

$\frac{1}{\sqrt{3} + \sqrt{2} } .\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2} } +\\\frac{2}{\sqrt{5} - \sqrt{3} } .\frac{\sqrt{5} + \sqrt{3} }{\sqrt{5} + \sqrt{3} } - \\\frac{3}{\sqrt{2} + \sqrt{5} } .\frac{\sqrt{2} - \sqrt{5} }{\sqrt{2} - \sqrt{5} }$

Doing the above operation we end up with :

$\frac{\sqrt{3} - \sqrt{2}}{ 3 - 2 } +\frac{2(\sqrt{5} + \sqrt{3}) }{5 - 3 } - \frac{3(\sqrt{2} - \sqrt{5} )}{2 - 5 }$

furthur simplifying:

= $\frac{\sqrt{3} - \sqrt{2}}{ 1 } +\frac{2(\sqrt{5} + \sqrt{3}) }{2 }+\frac{3(\sqrt{2} - \sqrt{5} )}{3}$

taking LCM

= $\frac{6(\sqrt{3} - \sqrt{2})+ 6(\sqrt{5} + \sqrt{3})+ 6(\sqrt{2} - \sqrt{5})}{ 6}$

= $\sqrt{3} - \sqrt{2}+\sqrt{5} + \sqrt{3}+ \sqrt{2} - \sqrt{5}$

=  2√3 [ Ans.]

5) $\frac{3 + \sqrt{2}}{ 3 -\sqrt{2}} = a + b \sqrt{2}$

Rationalizing

we get

= $\frac{9+ 2 + 6\sqrt{2} }{ 9 - 2}$

=  $\frac{11 + 6\sqrt{2} }{ 7}$

= $\frac{11}{7}$+$\frac{6\sqrt{2}}{7}$

⇒ a = $\frac{11}{7}$ and b =$\frac{6}{7}$  [ Ans.]

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Ps. Cheers ; p