Question

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

Answer 1

Answer:

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

hope this answer will help you :)