Solution for real \(a,b,c\) in
a⌊a⌋+c{c}−b{b}=0.16a\lfloor a \rfloor + c\{c\} - b\{b\} = 0.16a⌊a⌋+c{c}−b{b}=0.16
b⌊b⌋+a{a}−c{c}=0.25b\lfloor b \rfloor + a\{a\} - c\{c\} = 0.25b⌊b⌋+a{a}−c{c}=0.25
c⌊c⌋+b{b}−a{a}=0.49c\lfloor c \rfloor + b\{b\} - a\{a\} = 0.49c⌊c⌋+b{b}−a{a}=0.49
Where ⌊x⌋=\lfloor x \rfloor = ⌊x⌋= floor part of xxx
and {x}=\{x\} = {x}= fractional part of xxx.
Note by Jagdish Singh 7 years, 8 months ago
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