Math, asked by ranaayush473, 6 months ago

rationalise 3÷3√5+√2​

Answers

Answered by IIMidnightHunterII
3

Answer:

 =  \frac{3}{3 \sqrt{5} +  \sqrt{2}  }  \\  \\  =  \frac{3}{3 \sqrt{5} +  \sqrt{2}  }  \times  \frac{3 \sqrt{5}  -  \sqrt{2} }{3 \sqrt{5}  -  \sqrt{2} }  \\  \\ the \: denominator \: is \: given \: in \: the \: form \\ (a + b)(a - b) =  {a}^{2}  -  {b}^{2}  \\  \\ here \: a \:  = 3 \sqrt{5}  \\ b =  \sqrt{2}  \\  \\ so \\   =  \frac{3(3 \sqrt{5} -  \sqrt{2}  )}{(3  \sqrt{5}  )^{2} -  { \sqrt{2} }^{2}  }   \\  \\  =  \frac{9 \sqrt{5} - 3 \sqrt{2}  }{(9 \times 5) - 2}  \\  \\  =  \frac{9 \sqrt{5}  - 3 \sqrt{2} }{45 - 2}  \\  \\  =  \frac{9 \sqrt{5 }  - 3 \sqrt{2} }{43}  \\  \\  \\ this \: is \: the \: answer....

Answered by sweetkalpana14
0

Answer:

Answer:

\begin{gathered} = \frac{3}{3 \sqrt{5} + \sqrt{2} } \\ \\ = \frac{3}{3 \sqrt{5} + \sqrt{2} } \times \frac{3 \sqrt{5} - \sqrt{2} }{3 \sqrt{5} - \sqrt{2} } \\ \\ the \: denominator \: is \: given \: in \: the \: form \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ here \: a \: = 3 \sqrt{5} \\ b = \sqrt{2} \\ \\ so \\ = \frac{3(3 \sqrt{5} - \sqrt{2} )}{(3 \sqrt{5} )^{2} - { \sqrt{2} }^{2} } \\ \\ = \frac{9 \sqrt{5} - 3 \sqrt{2} }{(9 \times 5) - 2} \\ \\ = \frac{9 \sqrt{5} - 3 \sqrt{2} }{45 - 2} \\ \\ = \frac{9 \sqrt{5 } - 3 \sqrt{2} }{43} \\ \\ \\ this \: is \: the \: answer....\end{gathered}

=

3

5

+

2

3

=

3

5

+

2

3

×

3

5

2

3

5

2

thedenominatorisgivenintheform

(a+b)(a−b)=a

2

−b

2

herea=3

5

b=

2

so

=

(3

5

)

2

2

2

3(3

5

2

)

=

(9×5)−2

9

5

−3

2

=

45−2

9

5

−3

2

=

43

9

5

−3

2

thisistheanswer....

Similar questions