Question

$\frac{x {}^{2} + 9y {}^{2} + 6xy }{x - 3y}$
якщо x = 2016, а y = 1/3
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Answer 1

$\frac{x^{2}+9y^{2}+6xy }{x-3y}=\frac{(x+3y)^{2} }{x-3y} =\frac{(2016+3*\frac{1}{3})^{2} }{2016-3*\frac{1}{3} }=\frac{(2016+1)^{2} }{2016-1} =\frac{2017^{2} }{2015} =\frac{4068289}{2015}$