Đặt \(\left\{{}\begin{matrix}a+b=8+x\\b=3+y\end{matrix}\right.\left(x,y\in N,xy\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5+x-y\\b=3+y\end{matrix}\right.\)
Khi đó:
\(27a^2+10b^3=27\left(5+x-y\right)^2+10\left(3+y\right)^3\)
\(=27\left(25+x^2+y^2+10x-2xy-10y\right)+10\left(27+y^3+9y^2+27y\right)\)
\(=945+27\left(x^2+y^2-2xy\right)+270x+10y^3+90y^{2\text{}}\)
\(=945+27\left(x-y\right)^2+270x+10y^3+90y^2>945\)
Vậy \(27a^2+10b^3>945\)
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