Question

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

Answer 1

1/3-√8-1/√8-√7+1/√7-√6-1/√6-√5+1/√5-2=5
Now,
1/3-√8 = 3+√8,
1/√8-√7 = √8+√7,
1/√7-√6 = √7+√6,
1/√6-√5 = √6+√5,
1/√5-2 = √5+2
∴ 1/3-√8-1/√8-√7+1/√7-√6-1/√6-√5+1/√5-2
= 3+√8-√8-√7+√7+√6-√6-√5+√5+2
=3+2 = 5
Hence proved

Answer 2

Step-by-step explanation:

$\bf \underline{Given-}$

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

$\bf \underline{What \: to \: do-}$

1st we Rationalise all the denominator.

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

$\bf \underline{Solution-}$

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.

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

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

$⟹3 + 2=5$

$⟹5=5$

L.H.S = R.H.S.

Hence proved

I hope this helps..☺