Question

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

Answer 1

1(/3-√8)*(3+√8)/(3+√8)-1/(√8-√7)*(√8+√7)/(√8+√7)+1/(√7-√6)*(√7+√6)/(√7+√6)-1/(√6-√5)*(√6+√5)/(√6+√5)+1/(√5-2)*(√5+2)/(√5+2)
=(3+√8)/(3)²-(√8)²-(√8+√7)/(√8)²-(√7)²+(√7+√6)/(√7)²-(√6)²-(√6+√5)/(√6)²-(√5)²+(√5+2)/(√5)²-(2)²
=(3+√8)/1-(√8+√7)/1+(√7+√6)/1-(√6+√5)/1+(√5+2)/1
=3+√8-√8-√7+√7+√6-√6-√5+√5+2
=3+2
=5
hence proved
I hope this answer helps you

Answer 2

$\dfrac{1}{3-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-2}=5,$ proved.

Step-by-step explanation:

Show that, $\dfrac{1}{3-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-2}=5$

L.H.S.$=\dfrac{1}{3-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-2}$

Rationalising numerator and denominator, we get

$=\dfrac{1}{3-\sqrt{8}}\times \dfrac{3+\sqrt{8}}{3+\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{7}}\times\dfrac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{6}}\times \dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{5}}\times \dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}+\dfrac{1}{\sqrt{5}-2}\times \dfrac{\sqrt{5}+2}{\sqrt{5}+2}$

Using identity,

$a^{2}-b^{2}=(a+b+(a-b)$

Denominator part becomes 1,

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

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

= 3 + 2 = 5

= R.H.S. proved.