Math, asked by Tharani123, 1 year ago

plz send with full method

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Answers

Answered by Ronaldo1226
4
me kyu kru bta ....

tumhe method nahi aate iska matlab ye nahi ki sabko nahi aate
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Tharani123: because i am thinking you are elder than me
Tharani123: 16
Ronaldo1226: 23
Tharani123: now you tell
Tharani123: 23 is it true
Ronaldo1226: hmn
Ronaldo1226: ur clss?
Tharani123: by the way your answer is wrong
Ronaldo1226: to khud kr lo itna hi hai
Tharani123: ha to mene kud kiy lekin last step nahi a raha to mene pocha agar apko nahi ata tu plz fight mat karo
Answered by siddhartharao77
9
 \frac{1}{ \sqrt{3} -  \sqrt{2}  + 1}  *  \frac{ \sqrt{3}  +  \sqrt{2} - 1 }{ \sqrt{3}  +  \sqrt{2}  - 1}

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

 \frac{ \sqrt{3} +  \sqrt{2} - 1  }{ 3 +  \sqrt{6} -  \sqrt{3} -  \sqrt{6} - 2 +  \sqrt{2} +  \sqrt{3}  +  \sqrt{2} - 1 }

 \frac{ \sqrt{3} +  \sqrt{2} - 1  }{2 - 2 +  \sqrt{2} +  \sqrt{2}  }

 \frac{ \sqrt{3}  +  \sqrt{2} - 1 }{2 \sqrt{2} }

Multiply and divide by root2, we get

 \frac{ \sqrt{3} +  \sqrt{2} - 1  }{2 \sqrt{2} } *   \frac{ \sqrt{2} }{ \sqrt{2} }

 \frac{( \sqrt{3} +  \sqrt{2} - 1) *  \sqrt{2}   }{2 \sqrt{2}   * \sqrt{2} }

 \frac{ \sqrt{6} + 2 -  \sqrt{2}  }{4}


Hope this helps!


siddhartharao77: Any doubts... Ask me.. Gud luck!
Tharani123: my maths teacher said that we will take +1 as a lowest rationalising factor so we will do like this 1/√3-√2+1 × √3-√2-1/√3-√2-1
siddhartharao77: We can do in n number of ways.
Tharani123: ok thank you
siddhartharao77: Thanks Sis...
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