Kurosaki Akatsu Coordinator
09/03/2017 at 21:15-
Vũ Thị Hương Giang 09/03/2017 at 21:28a) 4xy3 + \(\dfrac{1}{3}\)xy8 + \(\dfrac{2}{5}\)xy3 + \(-\dfrac{1}{3}\)xy8
= 4xy3 + \(\dfrac{2}{5}\)xy3 + \(\dfrac{1}{3}xy^8\) + \(-\dfrac{1}{3}xy^8\)
= \(\dfrac{14}{3}xy^3+0=\dfrac{14}{3}xy^3\)
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a) \(4xy^3+\dfrac{1}{3}xy^8+\dfrac{2}{3}xy^3+\left(-\dfrac{1}{3}\right)xy^8\)
\(=\left(4+\dfrac{2}{3}\right)xy^3+\left[\left(-\dfrac{1}{3}\right)+\dfrac{1}{3}\right]xy^8\)
\(=\dfrac{14}{3}xy^3+0\)
\(=\dfrac{14}{3}.\dfrac{2}{3}.\left(-\dfrac{1}{2}\right)^3=-\dfrac{7}{18}\)
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FA FIFA Club World Cup 2018 16/01/2018 at 21:58a) 4xy3+13xy8+23xy3+(−13)xy8
=(4+23)xy3+[(−13)+13]xy8
=143xy3+0
=143.23.(−12)3=−71
