Math, asked by BKF, 1 year ago

rationalize the denominator of 6 upon root 5 + root 2​

Answers

Answered by Rathin5678
67

6/√5+√2

6/√5+√2×√5-√2/√5-√2

6(√5-√2)/5-2...(a+b)(a-b)=a^2-b^2

6(√5-√2)/3

2(√5-√2).

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Rathin5678: plz mark it as the Brainliest
Answered by payalchatterje
2

Answer:

After rationalization the given expression is 2( \sqrt{5}  -  \sqrt{2} )

Step-by-step explanation:

Given expression is 6 upon root 5 + root 2  =  \frac{6}{ \sqrt{5}  +  \sqrt{2} }

This is a problem of Power of indices.

Here we want to rationalize the denominator of given expression.

For rationalization we are multiplying denominator and numerator by ( \sqrt{5}  -  \sqrt{2} )and get,

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

After rationalization the given expression is

2 \times ( \sqrt{5}  -  \sqrt{2} )

Here applied formula,

 {x}^{2}  -  {y}^{2}  = (x + y)(x - y)

Some important formulas of Power of indices,

{x}^{0}  = 1 \\  {x}^{1}  = x \\  {x}^{a}  \times  {x}^{b}  =  {x}^{a + b}  \\  \frac{ {x}^{a} }{ {x}^{b} }  =  {x}^{a - b} \\  {x}^{ {y}^{a} }   =  {x}^{ya}  \\  {x}^{ - 1}  =  \frac{1}{x}  \\  {x}^{a}  \times  {y}^{a}  =  {(xy)}^{a}

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