Math, asked by ekanshrawat930, 19 days ago

show that : 1/3-√8 - 1/√8-√7 + 1/√7-√6 -1/√6-√5 + 1/√5-2 = 5

Answers

Answered by Salmonpanna2022
8

Step-by-step explanation:

Given:-

 \frac{1}{3 -  \sqrt{8} }  -  \frac{1}{ \sqrt{8} -  \sqrt{7}  }  +  \frac{1}{ \sqrt{7}  -  \sqrt{6} } -  \frac{1}{ \sqrt{6} -  \sqrt{5}}    + \frac{1}{ \sqrt{5} - 2 }  \\  \\

What to do:-

1st we Rationalise all the denominator.

2nd we arrange all according to the given question and simplify and last we get the answer.

Solution:-

We have,

 \frac{1}{3 -  \sqrt{8} }  -  \frac{1}{ \sqrt{8} -  \sqrt{7}  }  +  \frac{1}{ \sqrt{7}  -  \sqrt{6} } -  \frac{1}{ \sqrt{6} -  \sqrt{5}}    + \frac{1}{ \sqrt{5} - 2 }=5  \\  \\

Now,

Rationalising each term:

First \:  term: \:  \frac{1}{3 -  \sqrt{8} }  \\  \\

The denominator is 3-√8. Multiplying the numerator and denomination by 3+√8, we get

⟹ \frac{1}{3 -  \sqrt{8} }  \times  \frac{3  +   \sqrt{8} }{3 +  \sqrt{8} }  \\  \\

⟹ \frac{3 +  \sqrt{8} }{(3 -  \sqrt{8} )(3 +  \sqrt{8} )}  \\  \\

⬤ Applying Algebraic Identity

(a+b)(a-b) = a² - b² to the denominator

We get,

⟹ \frac{3 +  \sqrt{8} }{(3 {)}^{2} - ( \sqrt{8}   {)}^{2} }  \\  \\

⟹ \frac{3 +  \sqrt{8} }{9 - 8}  \\  \\

⟹ \frac{3 +  \sqrt{8} }{1}  \\  \\

⟹ \red{3 +  \sqrt{8} } \\  \\

Second  \: term: \:  \frac{1}{ \sqrt{8} -  \sqrt{7}  }  \\  \\

The denominator is √8-√7. Multiplying the numerator and denomination by √8+√7, we get

⟹ \frac{1}{ \sqrt{8}  -  \sqrt{7} }   \times  \frac{ \sqrt{8} +  \sqrt{7}  }{ \sqrt{8} +  \sqrt{7}  }  \\  \\

⟹ \frac{ \sqrt{8} +  \sqrt{7}  }{( \sqrt{8}  -  \sqrt{7})( \sqrt{8}   +  \sqrt{7} )}  \\  \\

⬤ Applying Algebraic Identity

(a+b)(a-b) = a² - b² to the denominator

We get,

⟹ \frac{ \sqrt{8}  +  \sqrt{7} }{( \sqrt{8} {)}^{2} - ( \sqrt{7}   {)}^{2}  }  \\  \\

⟹ \frac{ \sqrt{8} +  \sqrt{7}  }{8 - 7}  \\  \\

⟹ \frac{ \sqrt{8}  +  \sqrt{7} }{1}  \\  \\

⟹ \red{ \sqrt{8}  +  \sqrt{7} } \\  \\

Third  \: term  : \:  \:  \frac{1}{ \sqrt{7} -  \sqrt{6}  }  \\  \\

The denominator is √7-√6. Multiplying the numerator and denomination by √7+√6, we get

⟹ \frac{1}{ \sqrt{7} -  \sqrt{6}  }  \times   \frac{ \sqrt{7}  +  \sqrt{6} }{ \sqrt{7}  +  \sqrt{6} }  \\  \\

⟹ \frac{ \sqrt{7} +  \sqrt{6}  }{( \sqrt{7 }  -  \sqrt{6} )( \sqrt{7}  +  \sqrt{6}) }  \\  \\

⬤ Applying Algebraic Identity

(a+b)(a-b) = a² - b² to the denominator

We get,

⟹ \frac{ \sqrt{7}  +  \sqrt{6} }{( \sqrt{7} {)}^{2}  - ( \sqrt{6} {)}^{2}   }  \\  \\

⟹ \frac{ \sqrt{7} +  \sqrt{6}  }{7 - 6}  \\  \\

⟹ \frac{ \sqrt{7}  +  \sqrt{6} }{1}  \\  \\

⟹\red{ \sqrt{7}  +  \sqrt{6} } \\  \\

Fourth \:  term :  \:  \frac{1}{ \sqrt{6} -  \sqrt{5}  }  \\  \\

The denominator is √6-√5. Multiplying the numerator and denomination by √6+√5, we get

⟹ \frac{1}{ \sqrt{6}  -  \sqrt{5} }  \times  \frac{ \sqrt{6}  +  \sqrt{5} }{ \sqrt{6} +  \sqrt{5}  }  \\  \\

⟹ \frac{ \sqrt{6}  +  \sqrt{5} }{( \sqrt{6}  -  \sqrt{5})( \sqrt{6}   +  \sqrt{5} )}  \\  \\

⬤ Applying Algebraic Identity

(a+b)(a-b) = a² - b² to the denominator

We get,

⟹ \frac{ \sqrt{6} +  \sqrt{5}  }{( \sqrt{6} {)}^{2}  - ( \sqrt{5}  {)}^{2}  }  \\  \\

⟹ \frac{ \sqrt{6}  +  \sqrt{5} }{6 - 5}  \\  \\

⟹ \frac{ \sqrt{6} +  \sqrt{5}  }{1}  \\  \\

⟹  \red{\sqrt{3}  +  \sqrt{5} } \\  \\

Fifth  \: term: \:   \:  \: \frac{1}{ \sqrt{5} - 2 }  \\  \\

The denominator is √5-2. Multiplying the numerator and denomination by √5+2, we get

⟹ \frac{1}{ \sqrt{5}  - 2}  \times  \frac{ \sqrt{5} + 2 }{ \sqrt{5} + 2 }  \\  \\

⟹ \frac{ \sqrt{5} + 2 }{( \sqrt{5}  - 2)( \sqrt{5}  + 2)}  \\  \\

⬤ Applying Algebraic Identity

(a+b)(a-b) = a² - b² to the denominator

We get,

⟹ \frac{ \sqrt{5}  + 2}{( \sqrt{5} {)}^{2}  - ( 2 {)}^{2}  }  \\  \\

⟹ \frac{ \sqrt{5}+ 2}{5 - 4}  \\  \\

⟹ \frac{ \sqrt{5} + 2 }{1}  \\  \\

⟹ \red{ \sqrt{5}  + 2} \\  \\

Now, arranging all the rationalised denominator according to the given question and simplify that.

⟹ \red{(3 +  \sqrt{8} ) - ( \sqrt{8}  +  \sqrt{7} ) + ( \sqrt{7} +  \sqrt{6} ) - ( \sqrt{6}  +  \sqrt{5} ) + ( \sqrt{5}  + 2) }=5 \\  \\

⟹3 +  \cancel{ \sqrt{8} } -   \cancel{\sqrt{8}  }-   \cancel{\sqrt{7} } +   \cancel{\sqrt{7} } +   \cancel{\sqrt{6}  }-   \cancel{\sqrt{6}}   -   \cancel{\sqrt{5}}  +   \cancel{\sqrt{5}}  + 2 =5\\  \\

⟹3 + 2=5 \\  \\

⟹5=5 \\  \\

Hence, we showed that L.H.S = R.H.S.

  • I hope it's help you...☺
Similar questions