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Anil Gupta from DAV PUBLIC SCHOOL , Ludhiana
Subject :IMO    Class : Class 8



A a/b
B
C a2
D a - 2b 



Ans 1:

Asmi Raniga from Indian School Ghubra, Muscat
Class : Class 8
answer is told to be D not B;

Ans 2:

Samedh from Tree House High School, Pune
Class : Class 9
(a^2-b^2)/ab - (ab-b^2)/(ab- a^2) #Using special product = (a+b)(a-b)/ab - (ab-b^2)/(ab- a^2) #Removing out a and b common = (a+b)(a-b)/ab - b(a-b)/a(b-a) #LCM of both denominators i.e. ab and a(b-a) is ab(b-a) #Hence we multiply first equation by (b-a) and second equation by b = (a+b)(a-b)(b-a)/ab(b-a) - b^2(a-b)/ab(b-a) #Taking the denominator common = [(a+b)(a-b)(b-a) - b^2(a-b)]/ab(b-a) #Taking (b-a) common in the numerator =[(b-a)(a^2-b^2-b^2)]/ab(b-a) #Cancelling out (b-a) from numerator and denominator =a^2 - 2 b^2 / ab Hence B is the answer.

Ans 3:

Samedh from Tree House High School, Pune
Class : Class 9
(a^2-b^2)/ab - (ab-b^2)/(ab- a^2) #Using special product = (a+b)(a-b)/ab - (ab-b^2)/(ab- a^2) #Removing out a and b common = (a+b)(a-b)/ab - b(a-b)/a(b-a) #LCM of both denominators i.e. ab and a(b-a) is ab(b-a) #Hence we multiply first equation by (b-a) and second equation by b = (a+b)(a-b)(b-a)/ab(b-a) - b^2(a-b)/ab(b-a) #Taking the denominator common = [(a+b)(a-b)(b-a) - b^2(a-b)]/ab(b-a) #Taking (b-a) common in the numerator =[(b-a)(a^2-b^2-b^2)]/ab(b-a) #Cancelling out (b-a) from numerator and denominator =a^2 - 2 b^2 / ab Hence B is the answer.

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