Prove that if
acb^{2} bdc^{2}
 = 
a2bc b2c+d
then both of the fractions are also equal to
adbc

abc+d
(Assuming no divisions by zero, and b≠c.)
There is some error in the formulation of this. I offer an example that shows the error.
Suppose a=2 b=6 c=4 d=3
Then
(acb^2)/(a2bc) = (836)/(2124)=28/14=2
(bdc^2)/(b2c+d) = (1816)/(68+3)= 2/1=2
So, both fractions are equal, but:
(adbc)/(abc+d) = (624)/(264+3)= 18/5= 18/5
Differt from the two equal fractions.
This works only (with the example) if the numerator is acbd instead of adbc, or if the denominator of the third fraction is a instead of a.
Edited on February 2, 2018, 5:45 am

Posted by armando
on 20180201 17:06:58 